SIMOVERT MASTERDRIVES Application Manual SIMOVERT VC Cable drum Gearbox IM D Load Edition 09.99 09.99 No guarantee can be accepted for the general validity of the calculation techniques described and presented here. As previously, the configuring engineer is fully responsible for the results and the subsequently selected motors and drive converters. 2 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 Contents Contents 1 FOREWORD............................................................................................................ 7 2 FORMULAS AND EQUATIONS .............................................................................. 9 3 VARIOUS SPECIAL DRIVE TASKS ...................................................................... 25 3.1 3.1.1 3.1.2 3.1.3 3.1.4 3.1.5 3.1.6 3.1.7 Traction- and hoisting drives .................................................................................. 25 General information................................................................................................ 25 Traction drive with brake motor .............................................................................. 38 Hoisting drive with brake motor .............................................................................. 49 High-bay racking vehicle ........................................................................................ 60 Hoisting drive for a 20 t gantry crane...................................................................... 77 Elevator (lift) drive .................................................................................................. 83 Traction drive along an incline................................................................................ 92 3.2 3.2.1 3.2.2 3.2.3 3.2.4 3.2.5 Winder drives ....................................................................................................... 103 General information.............................................................................................. 103 Unwind stand with closed-loop tension control using a tension transducer........... 110 Winder with closed-loop tension control using a tension transducer..................... 117 Unwinder with 1FT6 motor ................................................................................... 123 Unwind stand with intermittent operation.............................................................. 131 3.3 3.3.1 3.3.2 3.3.3 Positioning drives ................................................................................................. 139 General information.............................................................................................. 139 Traction drive with open-loop controlled positioning using Beros ......................... 147 Elevator drive with closed-loop positioning control (direct approach).................... 149 3.4 3.4.1 3.4.2 3.4.3 3.4.4 3.4.5 Drives with periodic load changes and surge loads .............................................. 151 General information.............................................................................................. 151 Single-cylinder reciprocating compressor............................................................. 161 Three-cylinder pump ............................................................................................ 167 Transport system for freight wagons .................................................................... 171 Drive for an eccentric press.................................................................................. 184 3.5 3.5.1 3.5.2 3.5.3 Load distribution for mechanically-coupled drives ................................................ 197 General information.............................................................................................. 197 Group drive for a traction unit............................................................................... 204 Drive for an extraction tower with 10 mechanically-coupled motors...................... 207 3.6 Crank drive........................................................................................................... 208 3.7 Rotary table drive ................................................................................................. 221 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 3 Contents 09.99 3.8 Pivot drive ............................................................................................................ 227 3.9 Spindle drive with leadscrew ................................................................................ 238 3.10 Cross-cutter drive................................................................................................. 254 3.11 Centrifugal drive ................................................................................................... 261 3.12 Cross-cutter with variable cutting length............................................................... 272 3.13 Saw drive with crank ............................................................................................ 293 3.14 Saw drive as four-jointed system.......................................................................... 306 3.15 Mesh welding machine ......................................................................................... 323 3.16 Drive for a pantograph ......................................................................................... 357 3.17 Drive for a foil feed with sin2 rounding-off ............................................................. 377 4 INFORMATION FOR SPECIFIC APPLICATIONS ............................................... 387 4.1 4.1.1 4.1.2 4.1.3 4.1.4 4.1.5 Accelerating time for loads with square-law torque characteristics....................... 387 General information.............................................................................................. 387 Accelerating time for a fan drive........................................................................... 389 Accelerating time for a blower drive (using the field-weakening range) ................ 392 Re-accelerating time for a fan drive after a brief supply failure............................. 398 Calculating the minimum braking time for a fan drive ........................................... 406 4.2 4.2.1 4.2.2 Connecting higher output motors to a drive converter than is normally permissible.... 418 General information.............................................................................................. 418 Operating a 600 kW motor under no-load conditions ........................................... 422 4.3 4.3.1 4.3.2 4.3.3 Using a transformer at the drive converter output ................................................ 425 General information.............................................................................................. 425 Operating a 660 V pump motor through an isolating transformer ......................... 427 Operating 135V/200 Hz handheld grinding machines through an isolating transformer........................................................................................................... 428 4.4 4.4.1 Emergency off ...................................................................................................... 430 General information.............................................................................................. 430 4.5 Accelerating- and decelerating time with a constant load torque in the field-weakening range .......................................................................................... 436 General information.............................................................................................. 436 Braking time for a grinding wheel drive ................................................................ 438 4.5.1 4.5.2 4 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 Contents 4.6 Influence of a rounding-off function ...................................................................... 442 4.7 4.7.1 4.7.2 Possible braking torques due to system losses.................................................... 453 General information.............................................................................................. 453 Braking time for a fan drive .................................................................................. 460 4.8 4.8.1 4.8.2 Minimum acceleration time for fan drives with a linear acceleration characteristic........................................................................................................ 465 General information.............................................................................................. 465 Example ............................................................................................................... 471 4.9 4.9.1 4.9.2 Buffering multi-motor drives with kinetic energy ................................................... 474 General information.............................................................................................. 474 Example for buffering with an additional buffer drive ............................................ 479 4.10 Harmonics fed back into the supply in regenerative operation ............................. 483 4.11 4.11.1 4.11.2 4.11.3 Calculating the braking energy of a mechanical brake ......................................... 491 General information.............................................................................................. 491 Hoisting drive with counterweight ......................................................................... 491 Traversing drive ................................................................................................... 495 4.12 4.12.1 4.12.2 4.12.3 4.12.4 4.12.5 4.12.6 Criteria for selecting motors for clocked drives..................................................... 499 General information.............................................................................................. 499 Example 1, rotary table drive with i=4 and short no-load interval.......................... 507 Example 2, rotary table drive with i=4 and long no-load interval ........................... 510 Example 3, rotary table drive with a selectable ratio and short no-load interval.... 512 Example 4, rotary table drive with selectable ratio and long no-load interval ........ 515 Example 5, traversing drive with selectable ratio and longer no-load time............ 517 4.13 4.13.1 4.13.2 4.13.3 4.13.4 Optimum traversing characteristics regarding the maximum motor torque and RMS torque ................................................................................................... 520 Relationships for pure flywheel drives (high-inertia drives) ................................... 520 Relationships for flywheel drives (high-inertia drives) with a constant load torque 524 Summary.............................................................................................................. 526 Example with 1PA6 motor .................................................................................... 527 5 INDEX .................................................................................................................. 533 6 LITERATURE....................................................................................................... 536 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 5 Contents 6 09.99 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 1 1 Foreword Foreword This Application Manual for the new SIMOVERT MASTERDRIVES drive converter series is conceived as supplement to the Engineering Manual and the PFAD configuring/engineering program. Using numerous examples, it provides support when engineering complex drive tasks, such as for example traction- and hoisting drives, winders etc. It is shown how to calculate the torque- and motor outputs from the mechanical drive data and how to select the motor, drive converter, control type and additional components. As result of the diversity of the various drive tasks, it is not possible to provide detailed instructions with wiring diagrams, parameter settings etc. This would also go far beyond the scope of this particular document. In some cases, the examples are based on applications using previous drive converter systems and actual inquiries with the appropriate drive data. Further, this Application Manual provides information for specific applications, for example, using a transformer at the converter output, accelerating time for square-law load torque characteristics etc. This is supplemented by a summary of formulas and equations with units, formulas and equations for induction motors, converting linear into rotational motion, the use of gearboxes and spindles as well specifying and converting moments of inertia. Your support is required in updating the Application Manual, which means improvements, corrections, and the inclusion of new examples. This document can only become a real support tool when we receive the appropriate feedback as far as its contents are concerned and inquiries regarding the application in conjunction with the new SIMOVERT MASTERDRIVES drive converter series. This Application Manual is conceived as a working document for sales/marketing personnel. At the present time, it is not intended to be passed on to customers. Siemens AG SIMOVERT MASTERDRIVES - Application Manual 7 1 Foreword 8 09.99 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 2 Formulas and equations 2 Formulas and equations Units, conversion 1 N = 1 kgm/s2 = 0.102 kp force 1 kg mass 1W power 1 Ws = 1 Nm = 1 J work, energy 1 kgm2 = 1 Ws3 = 1 Nms2 moment of inertia g = 9.81 m/s2 acceleration due to the force of gravity Induction motor n0 = f 60 p synchronous speed f in Hz, n in RPM p: Pole pair number, f: Line supply frequency s= n0 - n n0 fs = n0 - n f n0 P = 3 U I cos = M= slip slip frequency M n 9550 9550 P n shaft output torque M in Nm, P in kW, n in RPM M n0 - n Mn n0 - nn torque in the range n0 to nn Mn: Rated torque, nn: Rated speed M stall ( V 2 ) M stall n Vn Siemens AG SIMOVERT MASTERDRIVES - Application Manual stall torque at reduced voltage 9 2 Formulas and equations M stall ( 09.99 fn 2 ) M stall n f stall torque in the field-weakening range Vn: Rated voltage, fn: Rated frequency Mstall n: Rated stall torque RMS torque The RMS torque can be calculated as follows if the load duty cycle is small with respect to the motor thermal time constant: M RMS = M 2 i ti i te + k f t p The following must be true: MRMS Mpermissible for S1 duty ti te tp kf time segments with constant torque Mi power-on duration no load time reduction factor (de-rating factor) Factor kf for the no-load time is a derating factor for self-ventilated motors. For 1LA5/1LA6 motors it is 0.33. For 1PQ6 motors, 1PA6 motors and 1FT6/1FK6 motors, this factor should be set to 1. Example M M RMS t t 1 t 10 2 t p t e Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 2 Formulas and equations Comparison of the parameters for linear- and rotational motion Linear motion Rotational motion a= dv dt [m/s2] acceleration = d dt [s-2] angular acceler. v= ds = a dt dt [m/s] velocity = d = dt [s-1] dt angular velocity s = v dt [m] distance = dt [rad] angle m [kg] mass J [kgm2] moment of inertia F [N] force M [Nm] torque F = m a [N] accelerating force M = J [Nm] acceler. torque P = F v [W] power P = M [W] power W = Fs [Ws] work W = M [Ws] work [Ws] kinetic energy W= [Ws] rotational energy W= 1 m v 2 2 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 1 J 2 2 11 2 Formulas and equations 09.99 Example : Relationships between acceleration, velocity and distance for linear motion Acceleration a Area corresponds to velocity v t Velocity v Area corresponds to distance s t Distance s t 12 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 2 Formulas and equations Converting linear- into rotational motion Example: Hoisting a load with constant velocity F s r , n v m v r velocity [m/s] roll radius [m] t time [s] angular velocity [s-1] speed [RPM] angle [rad] distance [m] m mass moved [kg] F = m g hoisting force [N] M = F r = m g r torque [Nm] P = M = F v = m g v W = F s = M = m g s power [W] hoisting work [Ws] v r v n= 60 = 60 2 2r = t s = vt = r = Siemens AG SIMOVERT MASTERDRIVES - Application Manual 13 2 Formulas and equations 09.99 Example: Accelerating a mass on a conveyor belt with constant acceleration a, v m s F r , a r acceleration [m/s2] roll radius [m] t accelerating time [s] angular acceleration [s-2] velocity [m/s] angular velocity [s-1] speed [RPM] angle [rad] distance [m] m mass moved [kg] F = m a accelerating force [N] M = F r = ma r accelerating torque [Nm] a = r v = a t = r t v = = t r v n= 60 = 60 2 2r t2 = 2 2 t v t s = a = 2 2 P = M = F v = ma v accelerating power 1 1 W = F s = M = m v 2 = m r 2 2 accelerating work 2 2 14 [W] [Ws] Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 2 Formulas and equations Converting linearly moved masses into moments of inertia The conversion is realized by equating the kinetic energy to the rotational energy, i. e. 1 1 m v2 = J 2 2 2 and J = m ( v 2 ) Example: Conveyor belt m v = r v r J = mr2 moment of inertia referred to the roll [kgm2] m r mass [kg] radius [m] Example: Hoisting drive r 2v = m J= m r 1 m r2 4 2v r v moment of inertia referred to the fixed roll [kgm2] mass [kg] fixed roll radius [m] Siemens AG SIMOVERT MASTERDRIVES - Application Manual 15 2 Formulas and equations 09.99 Using gearboxes i= n Motor nload gearbox ratio gearbox efficiency Referring moments of inertia to the motor shaft i, JMotor n Motor n Load J Load Gearbox J load = J load i2 J = J motor + J load i2 load moment of inertia referred to the motor shaft [kgm2] total moment of inertia referred to the motor shaft (without [kgm2] gearbox and coupling) Referring load torques to the motor shaft Example: Hoisting drive, hoisting and lowering with constant velocity i, r Motor Gearbox m 16 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 2 Formulas and equations r roll radius [m] FH = m g hoisting force [N] M load = m g r load torque at the roll [Nm] M load i load torque referred to the motor shaft when hoisting [Nm] M load i load torque referred to the motor shaft when lowering [Nm] (energy flow from the load to the motor) M load = M load = (energy flow from the motor to the load) Gearbox ratio for the shortest accelerating time with a constant motor torque a) Mload = 0, i. e., pure high-inertia drive without friction i= J load J motor and J load = Jmotor i2 this means, that the load moment of inertia, referred to the motor shaft, must be the same as the motor moment of inertia b) Mload 0 , i. e., the drive accelerates against frictional forces i= M load M J + ( load ) 2 + load M motor M motor Jmotor Siemens AG SIMOVERT MASTERDRIVES - Application Manual 17 2 Formulas and equations 09.99 Using spindle drives m Guide v, a D Motor M n h Sp motor l motor v feed velocity [m/s] hSp pitch [m] D spindle diameter [m] l spindle length [m] spindle moment of inertia (steel) [kgm2] moved masses [kg] motor speed [RPM] spindle pitch angle [rad] friction angle of the spindle [rad] J spindle D l 7.85 ( ) 4 1000 2 2 m nmotor = v 60 hSp SW = arctan = arctan hSp D hSp D - SW spindle efficiency 2 hSp angular acceleration - deceleration of the spindle [s-2] a b,v acceleration- deceleration of the load [m/s2] M b,v motor = J mot b,v Sp accelerating- and decelerating torque for the [Nm] b,v Sp = a b,v motor M b,v SP = J Sp b,v Sp accelerating- and decelerating torque for the [Nm] spindle Fb,v = m ab,v 18 accelerating- and deceleration force for mass m [N] Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 2 Formulas and equations Horizontal spindle drive FR = m g wF frictional force in the guide wF specific traversing resistance [N] Motor torque when accelerating M motor = M b motor + M b Sp + ( Fb + FR ) D tan( SW + ) 2 [Nm] Motor torque when decelerating M motor = - M v motor - M v Sp + ( - Fv + FR ) D tan( SW + sign( - Fv + FR )) 2 [Nm] Vertical spindle drive FH = m g hoisting force [N] Motor torque when accelerating upwards M motor = M b motor + M b Sp + ( Fb + FH ) D tan( SW + ) 2 [Nm] Motor torque when decelerating downwards M motor = - M v motor - M v Sp + ( - Fv + FH ) D tan( SW + sign( - Fv + FR )) 2 [Nm] Motor torque when accelerating upwards M motor = - M b motor - M b Sp + ( - Fb + FH ) D tan( SW - sign( - Fb + FH )) 2 [Nm] Motor torque when decelerating downwards M motor = M v motor + M v Sp + ( Fv + FH ) D tan( SW - ) 2 [Nm] Blocking occurred for a downwards movement for > SW Siemens AG SIMOVERT MASTERDRIVES - Application Manual 19 2 Formulas and equations 09.99 Accelerating- and braking time at constant motor torque and constant load torque ta = tbr = 2 n max J ( ) accelerating time from 0 to nmax [s] ) braking time from nmax to 0 [s] 60 M motor - M load 2 nmax J 60 M motor + M load ( J in kgm2, M in Nm, nmax in RPM (final motor speed) (moment of inertia and load torques converted to the motor shaft) Accelerating time at constant motor torque and for a square-law load torque Example: Fan M load = M load max ( ta = n nmax )2 nmax J ln 60 M motor M load max M motor +1 M load max M motor -1 M load max accelerating time from 0 to nmax [s] J in kgm2, M in Nm, nmax in RPM (final motor speed) Mmotor > Mload max when accelerating, otherwise the final speed will not be reached. 20 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 2 Formulas and equations Moments of inertia of various bodies (shapes) Solid cylinder r l J= 1 m r 2 = l r 4 103 2 2 J= 1 m ( R 2 + r 2 ) = l ( R 4 - r 4 ) 103 2 2 Hollow cylinder R r l Thin-walled hollow cylinder rm l J = m rm2 = 2 l rm3 103 (r R rm ) J in kgm2 l, r, R, rm, in m m in kg in kg/dm3 (e. g. steel: 7.85 kg/dm3) Siemens AG SIMOVERT MASTERDRIVES - Application Manual 21 2 Formulas and equations 09.99 Calculating the moment of inertia of a body If the moment of inertia of a body around the center of gravity S is known, then the moment of inertia to a parallel axis A is: m S s A J = JS + m s2 Example: Disk, thickness d with 4 holes J = J disk without hole - 4 ( J S hole + mhole s 2 ) S s r A = 10 3 d R 4 - 4 ( d r 4 + r 2 d s 2 ) 2 2 R J in kgm2 d, r, R, s in m m in kg in kg/dm3 22 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 2 Formulas and equations Traction drives - friction components For traction drives, the friction consists of the components for rolling friction, bearing friction and wheel flange friction. The following is valid for the frictional force (resistance force): FW = m g wF resistance force [N] m mass of the traction drive [kg] wF specific traction resistance If factor wF is not known, it can be calculated as follows: wF = 2 DW ( r + f ) + c D 2 D wheel diameter [m] DW shaft diameter for bearing friction [m] r bearing friction coefficient f lever arm of the rolling friction c coefficient for wheel flange friction [m] Values for wF can be taken from the appropriate equations/formulas (e. g. /5/, /6/). m. g D DW f Siemens AG SIMOVERT MASTERDRIVES - Application Manual 23 2 Formulas and equations 24 09.99 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks 3 Various special drive tasks 3.1 Traction- and hoisting drives 3.1.1 General information Traction drives Motor v D Basic traction drive Travelling at constant velocity FW = m g wF drag force (this always acts against the direction of [N] motion, i. e. it has a braking effect) m mass being moved wF specific traction resistance (this takes into account [kg] the influence of the rolling- and bearing friction) M load = FW D 2 D Pmotor = FW vmax 103 M motor = M load i load torque at the drive wheel [Nm] drive wheel diameter [m] motor output (power) at maximum velocity [kW] full-load, steady-state output motor torque Siemens AG SIMOVERT MASTERDRIVES - Application Manual [Nm] 25 3 Various special drive tasks 09.99 vmax max. velocity mech. efficiency i= nmotor nwheel nmotor max = i [m/s] gearbox ratio vmax 60 D motor speed at vmax [RPM] Note: For traction units used outside, in addition to resistance force FW , there is also the wind force (wind resistance) FWi. FWi = AWi pWi wind resistance (wind force) [N] AWi surface exposed to the force of the wind [m2] pWi wind pressure [N/m2] angular motor acceleration and -braking of the motor [s-2] ab,v max max. acceleration and braking of the traction unit [m/s2] M b ,v motor = J motor b ,v motor accelerating- and braking torque for the motor [Nm] J motor motor moment of inertia [kgm2] accelerating- and braking torque for the load, [Nm] Accelerating and braking b ,v motor = i ab ,v max Mb ,v load = J load D J load = m ( ) 2 2 26 2 D b ,v motor i referred to the drive wheel moment of inertia of the linearly-moved masses, [kgm2] referred to the drive wheel Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks The component of other moments of inertia such as couplings, gearboxes, drive wheels etc., is generally low, and can be neglected with respect to the load moment of inertia and the motor moment of inertia. Often, the rotating masses are taken into account by adding between 10 and 20 % to the linearly-moving masses. Motor torque when accelerating (starting torque): M motor = Mb motor + ( Mb load + M load ) 1 i [Nm] Motor torque when decelerating (braking): M motor = - M v motor + ( - M v load + M load )1) i (regenerative operation) [Nm] 1) If the expression in brackets should be > 0 for very low deceleration values, then the factor must be changed to 1/ (so that the deceleration component is not predominant). When accelerating (starting) the drive wheels should not spin. Force which can be transferred through a drive wheel: Fwheel FR FR = Fwheel frictional force due to static friction (stiction) [N] Fwheel force due to the weight acting on the drive wheel [N] coefficient of friction for the static friction, e. g. 0.15 for steel on steel Siemens AG SIMOVERT MASTERDRIVES - Application Manual 27 3 Various special drive tasks 09.99 The following must be valid for the rotating masses if the accelerating forces are neglected: F R > Fb Fb = m ab max accelerating force for linear motion [N] The maximum permissible acceleration is obtained as follows if all of the wheels are driven: ab max < g [m/s2] Example of a traction cycle Forwards v Interval Reverse Interval vmax t tb tk tv tp - vmax M motor t Pmotor t Regenerative operation 28 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks Hoisting drives Cable drum D Motor Gearbox Load Principle of operation of a hoisting drive (single-cable) Hoisting a load with constant velocity FH = m g hoisting force [N] m load [kg] load torque at the cable drum (it always acts in the [Nm] M load = FH D 2 D Pmotor = FH vmax 103 M load i same direction) cable drum diameter [m] motor output at the maximum hoisting velocity [kW] (steady-state full-load output) motor torque [Nm] vmax max. hoisting velocity [m/s] mechanical efficiency M motor = i= nmotor ndrum gearbox ratio Siemens AG SIMOVERT MASTERDRIVES - Application Manual 29 3 Various special drive tasks nmotor max = i vmax 60 D 09.99 motor speed at vmax [RPM] Lowering the load at constant velocity Pmotor = - FH vmax 103 motor output (regenerative operation) [kW] M motor = M load i motor torque [Nm] angular acceleration and angular deceleration of the [s-2] Acceleration and deceleration b ,v motor = i ab ,v max 2 D motor ab,v max max. acceleration or deceleration of the load [m/s2] M b ,v motor = J motor b ,v motor accelerating- or decelerating torque for the motor [Nm] J motor motor moment of inertia [kgm2] accelerating- or decelerating torque for the load [Nm] Mb , v load = J load D J load = m ( ) 2 2 b , v motor i referred to the cable drum load moment of inertia referred to the cable drum [kgm2] Hoisting the load, motor torque when accelerating (starting torque): M motor = Mb motor + ( Mb load + M load ) 1 i [Nm] Hoisting the load, motor torque when decelerating: M motor = - M v motor + ( - M v load + M load ) 30 1 i [Nm] Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks Lowering the load, motor torque when accelerating: M motor = - Mb motor + ( - Mb load + M load ) i (regenerative operation) [Nm] (regenerative operation) [Nm] Lowering the load, motor torque when decelerating: M motor = M v motor + ( M v load + M load ) i For hoisting drives, the influence of the masses to be accelerated (load and rotating masses) is generally low with respect to the load torque. Often, these are taken into account by adding 10% to the load torque. Example of a hoisting and lowering cycle Hoisting v Interval Lowering Interval vmax t tb tk tv tp - vmax M Motor t PMotor t Regen. op. Siemens AG SIMOVERT MASTERDRIVES - Application Manual 31 3 Various special drive tasks 09.99 Selecting an induction motor For traction- and hoisting drives, when selecting the induction motor, intermittent duty S3 can be applied (power-on duration 20 to 60%). Depending on the relative power-on duration and motor type, the thermally permissible motor output is increased according to the following formula: Pth Ppermissible 1 + (1 - t r ) h (for load duty cycles < 10 min; h, k , refer to Catalog 0 (1 - k 0 ) t r [kW] permissible output at the drive converter for S1 duty [kW] M11) Ppermissible (control range 1:2) tr = PON 100% PON = power-on duration Thus, for example, for a 4-pole 1LA6 motor, frame size 180 and 40% power-on duration, the output can be increased by approx. 25% referred to Ppermissible. This value is used as basis for the steady-state full-load output. When utilizing motors according to temperature rise class F, the rated output can generally be used for Ppermissible. For traction drives with high accelerating torques, the maximum torque is decisive when selecting the motor. It must have an adequate safety margin to the motor stall torque, and at least: M stall > 13 . M max Thus, even for line supply undervoltage conditions, safe, reliable operation is guaranteed. In order to guarantee the correct functioning of the closed-loop control, 200% of the rated motor torque should not be exceeded, i. e.: M max 2 M n (for SIMOVERT VC) If acceleration and braking have a significant influence on the load duty cycle, the RMS motor torque should also be calculated. The following must be true: M RMS = M perm. M 2 i ti te + k f t p M perm. (for load duty cycles < 10 min) permissible torque demanded from the drive [Nm] [Nm] converter for S1 duty (e.g. control range 1:2) ti time segments with constant torque Mi [s] te power-on duration [s] 32 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks kf derating factor tp no-load time [s] Factor kf for the no-load time is a derating factor for self-ventilated motors. For 1LA5/1LA6 motors it is 0.33. This factor should be set to 1 for 1PQ6 motors, 1PA6 motors and 1FT6/1FK6 motors. If the proportion of low speed duty within the load duty cycle cannot be neglected (e. g. a high proportion of time is taken for accelerating and decelerating or longer periods of duty at low speeds), then this must be appropriately taken into account. In this case, the arithmetic average of the speeds within the cycle time T is generated, so that the appropriate reduction factor can be taken from the reduction characteristic for S1 duty. nmot average = nmot i A + nmot i E 2 te + k f t P i [RPM] ti average motor speed in time sector ti (A: Initial value, [RPM] nmot i A + nmot i E I: Final value) 2 It is possible to also calculate the motor RMS current for a load duty cycle instead of MRMS for a more accurate analysis and for operation in the field-weakening range. The following must be true: ( I RMS = I mot i A + I mot i E i I mot i A + I mot i E 2 I perm. 2 te + k f t P )2 ti I perm. average motor current in time sector ti (A: Initial [A] [A] value, I: Final value) permissible current at the drive converter for S1 duty [A] at naverage (this is derived from the Mper S1 curve; for example, for 1PA6 motors, Iper=Imot n) Refer to the next Section Selecting the drive converter" to calculate the motor current. Siemens AG SIMOVERT MASTERDRIVES - Application Manual 33 3 Various special drive tasks 09.99 Selecting the drive converter Traction drives For acceleration and braking, the drive converter overload capability can be used for 60 s, as the accelerating- and braking times are generally low. The induction motor current at the maximum motor torque is approximately given by: I motor max ( M motor max M motor n I n 1 k n2 [A] rated motor magnetization current kn = 1 kn = 2 2 2 2 )2 ( I motor n - I n ) kn + I n n mot nmot n for nmot nmot n constant flux range for nmot > nmot n field-weakening range [A] From approx. Mmot 1.5 Mmot n, saturation effects occur in the motor current which can no longer be neglected. This can be taken into account by linearly increasing the motor current in the range from Mmot n to 2 Mmot n (0% at Mmot n, approx. 10% at 2 Mmot n). When selecting the drive converter, in addition to the maximum motor current, the RMS motor current, referred to the load duty cycle, must also be taken into account (in this case, kf must always be set to 1). The rated drive converter current must be greater than or equal to the RMS motor current within a 300 s interval. For a motor matched to the drive converter, this condition is generally fulfilled. It may be necessary to make a check, for example, for induction motors with high magnetizing currents and short no-load intervals. For traction drives, vector control is preferable due to the improved control characteristics (e. g. accelerating along the current limit) and also due to the improved immunity to stalling. Generally, vector control is also possible, even when several motors are connected to a single drive converter, as normally, the motors are similarly loaded. Hoisting drives For hoisting drives, the maximum induction motor current is calculated the same way as for traction drives. The proportion of accelerating- and decelerating torques in the load torque is generally low. Generally, the maximum motor current is only approximately 100 to 120 % of the rated motor current. As hoisting drives must be able to accelerate suspended loads, it is recommended that a higher safety margin is used between the calculated maximum motor current and the maximum drive converter current. Depending on the accelerating time, the starting current 34 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks should be approximately 150 to 200 % of the required motor current. Vector control is also preferable for hoisting drives due to the improved control characteristics, and as it is possible to impress a defined current when starting. A speed encoder improves the dynamic characteristics. Further, the tachometer can be used to monitor the setpoint- actual values (to identify if the load has dropped) . Generally, holding brakes are used to hold a suspended load. In order that the load doesn't drop when the brake is released, the motor must first develop an opposing torque to the brake. This means: * An appropriately high motor current setpoint is entered for closed-loop frequency control (depending on the starting torque) * The motor flux has been established (approx. 0.1 to 1 s depending on the motor size) * A slip frequency is specified for closed-loop frequency control * Speed controller pre-control for closed-loop speed control if the load is to be held at zero speed For closed-loop frequency control, after the signal to release the brake has been output, a short delay should be inserted before starting. This prevents the drive starting, in the open-loop controlled range, against a brake which still hasn't been completely released. After the load has been hoisted, the signal to close the brake is output at fU=0. The inverter is only inhibited somewhat later when it is ensured that the brake has been applied. During this delay time, the motor develops a holding torque as result of the DC current braking as f U=0. fU ~ ~ fslip t Start of acceleration Inverter inhibit Brake release signal Enable inverter, establish motor flux f U = 0, brake application signal Example for inverter enable, brake control and acceleration enable for a hoisting drive with closedloop frequency control. Siemens AG SIMOVERT MASTERDRIVES - Application Manual 35 3 Various special drive tasks 09.99 Dimensioning the braking resistor When dimensioning the brake resistor, the motor output in the regenerative mode as well as the load duty cycle are important. The maximum output in regenerative operation occurs when the drive is braked at maximum speed. Thus, the max. braking power of the brake resistor is given by: Pbr W max = Pbr motor max motor = M motor v nmotor max motor 9550 motor [kW] motor efficiency The maximum braking power of the brake resistor may not exceed the permissible peak braking power. Further, the following must also be true: Wbr Pbrcont . T T Wbr = Pbr W dt braking energy for one cycle [kWs] cycle time ( 90 s) [s] permissible continuous braking power [kW] 0 T Pbr cont . For cycle times greater than 90 s, for the condition Wbr Pbr cont . T a time segment T = 90 s is selected from the load duty cycle so that the highest average braking power occurs in that segment. For traction drives, the following is obtained, for example, when braking from vmax to 0 within time tv: Wbr = Pbr W max t v 2 [kWs] For hoisting drives, the braking power is constant when the load is lowered at constant velocity vmax: Pbr W const = 36 FH vmax motor 103 [kW] Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks Thus, the braking power during lowering is approximately given by: 1 1 Wbr Pbr W const ( tb + t k + t v ) 2 2 [kWs] tb accelerating time [s] tk lowering time at constant velocity vmax [s] tv decelerating time [s] Siemens AG SIMOVERT MASTERDRIVES - Application Manual 37 3 Various special drive tasks 3.1.2 09.99 Traction drive with brake motor This application involves a traction drive with brake motor (sliding-rotor motor). The motor is operated at a crawl speed for positioning. When the limit switch is reached, the motor is shutdown, and is mechanically braked down to standstill using the integral brake. Drive data Mass to be moved m = 4200 kg Load wheel diameter D = 0.27 m Specific traction resistance wF = 0.03 Gearbox ratio i = 18.2 Mech. efficiency = 0.85 Max. traversing speed vmax = 1.1 m/s Min. traversing velocity (crawl velocity) vmin = 0.1 m/s Max. acceleration ab max = 0.35 m/s2 Max. deceleration av max = 0.5 m/s2 Rated motor output Pn Rated motor current Imotor n = 6.2 A Rated motor torque Mn = 15.4 Nm Rated motor speed nn = 1425 RPM Motor efficiency motor = 0.79 Motor moment of inertia Jmotor = 0.012 kgm2 Coupling moment of inertia JK = 0.00028 kgm2 Max. distance moved smax = 3.377 m Cycle time tcycle = 60 s Crawl operation time ts = 1.2 s Brake application time tE = 0.1 s Mechanical brake torque Mmech = 38 Nm 38 = 2.3 kW Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks Constant velocity motion Resistance force, load torque: FW = m g wF = 4200 9.81 0.03 = 12361 . N M load = FW D 2 = 12361 . 0.27 = 166.9 Nm 2 Motor output, motor torque: Pmotor = FW vmax 12361 . 11 . = = 16 . kW 3 3 10 0.85 10 M motor = M load 166.9 = = 10.8 Nm i 18.2 0.85 Motor speed at vmax: nmotor max = i nload max = i vmax 60 . 60 11 = 18.2 = 1416 RPM D 0.27 Motor speed at vmin: nmotor min = i nload min = i 01 vmin 60 . 60 = 18.2 = 128.7 RPM D 0.27 Calculating the motor torques when accelerating and decelerating Angular acceleration and angular deceleration of the motor: b motor = i ab max 2 2 = 18.2 0.35 = 47.2 s -2 D 0.27 v motor = i av max 2 2 = 18.2 0.5 = 67.4 s -2 D 0.27 Accelerating torque and decelerating torque for motor and coupling: Mb motor + K = ( Jmotor + J K ) b motor = (0.012 + 0.00028) 47.2 = 0.58 Nm Siemens AG SIMOVERT MASTERDRIVES - Application Manual 39 3 Various special drive tasks M v motor + K 09.99 = ( Jmotor + J K ) v motor = (0.012 + 0.00028) 67.4 = 0.828 Nm Load moment of inertia referred to the load wheel: D 2 ) 2 0,27 2 = 4200 ( ) = 76.55 kgm2 2 J load = m ( Accelerating torque and decelerating torque for the load: Mb load = J load b motor 47.2 = 76.55 = 198.5 Nm 18.2 i M v load = J load v motor 67.4 = 76.55 = 2835 . Nm i 18.2 Motor torque when accelerating: M motor = Mb motor + K + ( M b load + M load ) = 0.58 + (198.5 + 166.9) 1 i 1 = 24.2 Nm 18.2 0.85 Motor torque when decelerating: M motor = - M v motor + K + ( - M v load + M load ) = -0.828 + ( -2835 . + 166.9) i 0.85 = -6.27 Nm 18.2 (regenerative operation) The highest motor torque is required when accelerating, namely 157 % of the motor rated torque. The moments of inertia for the gearbox and other rotating masses have been neglected. Accelerating time: tb = 40 vmax 11 . = = 314 . s ab max 0.35 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks Calculating the maximum braking power The maximum motor output in the regenerative mode occurs when the drive starts to decelerate from the maximum motor speed. Pbr motor max = M motor v nmotor max 9550 =- 6.27 1416 = -0.93 kW 9550 Drive converter selection For the highest required motor torque, a motor current of approximately: I motor max ( M motor max M motor n 2 2 2 ) 2 ( I motor n - I n ) + I n is obtained. The following is obtained with I n = 0.7 I motor n (assumption): I motor max ( 24.2 2 ) (6.22 - 0.72 6.2 2 ) + 0.7 2 6.2 2 = 8.2 A for tb=3.14 s 15.4 Selected drive converter: 6SE7018-0EA61 PV n=3 kW; IV n=8 A, IV max=11 A Closed-loop frequency control type Siemens AG SIMOVERT MASTERDRIVES - Application Manual 41 3 Various special drive tasks 09.99 Dimensioning the brake resistor The brake resistor is used during deceleration (braking). As the braking characteristics are the same for forwards as well as reverse motion, only forwards motion must be analyzed. Velocity-time diagram for forwards motion v Forwards motion No-load interval v max Motor powered-down Brake applied ~ ~ v min tb t ts tv tk tE t v mech T Maximum distance moved: smax = 3.377 m Cycle time for forwards- and reverse motion: T= t cycle = 30 s 2 Time for operation at the crawl speed: t s = 12 . s 42 (specified) Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks Accelerating time from 0 to vmax: tb = 314 . s Decelerating time from vmax to vmin: tv = vmax - vmin 11 . - 01 . = =2s av max 0.5 When neglecting the distance travelled during the brake application time tE and the deceleration time as a result of the mechanical brake tv mech, smax is given by the area in the v-t diagram: smax vmax tb v +v + vmax t k + max min t v + vmin t s 2 2 The travelling time at constant velocity vmax is given by: tk smax - vmax tb vmax + vmin - t v - vmin t s 2 2 vmax 3.377 - 11 . 3143 . 11 . + 01 . - 2 - 01 . 12 . 2 2 = 0.3 s 11 . No-load time: t p = T - t total = T - (tb + t k + t v + t s ) = 30 - (314 . + 0.3 + 2 + 12 . ) = 23.4 s Max. braking power of the brake resistor when decelerating from vmax to vmin: Pbr W max = Pbr motor max motor = 0.93 0.79 = 0.73 kW Min. braking power of the brake resistor when decelerating from vmax to vmin: Pbr W min = Pbr motor min motor = = M motor v nmotor min 9550 motor 6.27 128.7 0.79 = 0.067 kW 9550 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 43 3 Various special drive tasks 09.99 Braking diagram P br W Forwards motion No-load interval 0.73 kW Deceleration ~ ~ 0.067 kW 3.14 s t 2s 0.3 s 30 s Braking energy for one cycle (corresponds to the area in the braking diagram): Wbr = Pbr W max + Pbr W min 2 tv = 0.73 + 0.067 2 = 0.8 kWs 2 The following must be valid for the brake resistor: Wbr 0.8 = = 0.027 kW Pbr cont . 30 T With Pbr cont . = P20 36 (with an internal brake resistor) the following is obtained 36 0.027 = 0.972 kW P20 Thus, the smallest braking unit is selected with P 20 = 5 kW (6SE7018-0ES87-2DA0). The internal braking resistor is adequate. 44 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks Thermal motor analysis Torque characteristics for forwards motion M motor Forwards motion No-load interval 24.2 Nm 10.8 Nm 10.8 Nm ~ ~ 2s 3.14 s 1.2 s 23.4 s t 0.3 s -6.27 Nm 30 s The RMS torque is obtained from the torque characteristics: M RMS = = . + 10.82 0.3 + 6.27 2 2 + 10.82 12 . 24.2 2 314 . + 0.3 + 2 + 12 . + 0.33 23.4 314 2092.5 14.4 = 12.05 Nm Siemens AG SIMOVERT MASTERDRIVES - Application Manual 45 3 Various special drive tasks 09.99 Calculating the average speed nmotor , average = nmot i A + nmot i E ti 2 te + k f t P n + n min 1 nmax t b + n max t k + max t v + nmin t s 2 = 2 tb + t k + tv + t s + k f t P 1 1416 + 128.7 . + 1416 0.3 + 1416 314 2 + 128.7 1,2 2 = 2 = 301 RPM 314 . + 0.3 + 2 + 12 . + 0.33 23.4 For a reduction factor of 0.8 (control range 1:5), a permissible torque is obtained as follows: 0.8 M mot n = 0.8 15.4 = 12.32 Nm Operation is thermally permissible. Calculating the distance travelled after the brake motor has been powered-down When the traction unit reaches the limit switch at the crawl speed, the brake motor is powereddown by inhibiting the drive converter pulses. The brake application delay time t E now starts before the mechanical brake starts to become effective. The traction unit continues to move without any motor torque and is only braked by the effect of drag and the mechanical efficiency. After the brake is applied, it takes time tv mech to reach standstill. Motor shutdown v vmin Brake applied vE sv E ~ ~ s t tE t v mech 46 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks To calculate the distance travelled after the brake motor has been shutdown, a sub-division is made into range 1 (deceleration due to drag and the mechanical efficiency) and range 2 (deceleration as a result of the mechanical brake). Calculating the deceleration after shutdown (range 1) The following is obtained with Mmotor=0: M motor = 0 = - M v motor + K + ( - M v load + M load ) i or 0 = - ( J motor + J K ) v1 motor - J load v1 motor + M load 2 i i The motor angular deceleration is then given by: v1 motor = M load i J motor + J K + J load 2 i = 166.9 0.85 18.2 0.85 0.012 + 0.00028 + 76.55 18.2 2 = 37.35 s -2 and for the deceleration of the traction unit: a v1 = v1 motor D 37.35 0.27 = = 0.277 m / s 2 i 2 18.2 2 Distance travelled during tE: sE 1 = vmin t E - av1 t E2 2 1 = 01 . 01 . - 0.277 01 . 2 = 0.0086 m 2 Velocity after tE: v max E = v min - a v1 t E = 01 . - 0.277 01 . = 0.0723 m / s Calculating the deceleration as result of the mechanical brake (range 2) The following is obtained with Mmotor=-Mmech: M motor = - M mech = - M v motor + K + ( - M v load + M load ) Siemens AG SIMOVERT MASTERDRIVES - Application Manual i 47 3 Various special drive tasks 09.99 and - M mech = - ( J motor + J K ) v 2 motor - J load v 2 motor + M load 2 i i The motor angular deceleration is then given by: v 2 motor = M mech + M load i J motor + J K + J load i2 = 38 + 166.9 0.85 18.2 0.012 + 0.00028 + 76.55 0.85 18.2 2 = 219.4 s -2 and for the traction unit deceleration: av2 = v 2 motor D 219.4 0.27 . = = 1627 m / s2 i 2 18.2 2 Deceleration time as the result of the mechanical brake: t v mech = vE 0.0723 = = 0.044 s 1627 . av2 Distance travelled during deceleration caused by the mechanical brake: sv = v E t v mech 2 = 0.0723 0.044 = 0.0016 m 2 As the mechanical brake is only applied after a delay time tE, the traction unit travels a total distance of s ges = s E + sv = 0.0086 + 0.0016 = 0.01 m = 10 mm This load-dependent distance must be taken into account by the positioning control. 48 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3.1.3 3 Various special drive tasks Hoisting drive with brake motor This application involves a hoisting drive with brake motor (shift-rotor motor). Positioning is realized at the crawl speed. Drive data Mass to be moved mL = 146 kg Intrinsic mass mE = 66 kg Cable drum diameter D = 0.1 m Gearbox ratio i = 20.5 Mech. efficiency = 0.85 Max. hoisting velocity vmax = 0.348 m/s Min. hoisting velocity (crawl velocity) vmin = 0.025 m/s Max. acceleration amax = 0.58 m/s2 Rated motor output Pn = 1.05 kW Rated motor current Imotor n = 3.8 A Rated motor torque Mn = 7.35 Nm Rated motor speed nn = 1365 RPM Motor efficiency motor = 0.75 Motor moment of inertia Jmotor = 0.0035 kgm2 Coupling moment of inertia JK = 0.00028 kgm2 Max. hoisting height hmax = 0.8 m Cycle time tcycle = 90 s Crawl time ts = 0.4 s Brake application time tE = 0.04 s Mechanical brake torque Mmech = 14.5 Nm Siemens AG SIMOVERT MASTERDRIVES - Application Manual 49 3 Various special drive tasks 09.99 Hoisting the load with constant velocity Hoisting force, load torque: FH = ( mL + mE ) g = (146 + 66) 9.81 = 2079.7 N M load = FH D 2 = 2079.7 01 . = 104 Nm 2 Motor output, motor torque: Pmotor = FH v max 2079.7 0.348 = = 0.852 kW 10 3 0.85 10 3 M motor = M load 104 = = 5.97 Nm i 20.5 0.85 Motor speed at vmax: nmotor max = i nload max = i v max 60 0.348 60 = 20.5 = 1362 RPM . D 01 Motor speed at vmin: nmotor min = i nload min = i v min 60 0.025 60 = 20.5 = 97.9 RPM . D 01 Lowering the load at constant velocity Motor output, motor torque: Pmotor = - FH v max 2079.7 0.348 = - 0.85 = -0.615 kW 3 10 103 M motor = M load 104 = 0.85 = 4.31 Nm i 205 50 (regenerative operation) Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks Calculating the motor torques when accelerating and decelerating Motor angular acceleration: motor = i a max 2 2 = 20.5 0.58 = 237.8 s -2 D . 01 Accelerating torque for motor and coupling: M b motor + K = ( J motor + J K ) motor = (0.0035 + 0.00028) 237.8 = 0.899 Nm Load moment of inertia referred to the cable drum: D 2 ) 2 01 . = (146 + 66) ( ) 2 = 0.53 kgm2 2 J load = ( mL + mE ) ( Load accelerating torque: M b load = J load = 0.53 motor i 237.8 = 615 . Nm 20.5 As deceleration is the same as acceleration, the following is true: M v motor + K = M b motor + K M v load = M b load Hoisting the load, motor torque when accelerating: M motor = M b motor + K + ( M b load + M load ) = 0.899 + (615 . + 104) 1 i 1 = 7.22 Nm 20.5 0.85 Hoisting the load, motor torque when decelerating: M motor = - M v motor + K + ( - M v load + M load ) Siemens AG SIMOVERT MASTERDRIVES - Application Manual 1 i 51 3 Various special drive tasks 09.99 M motor = -0.899 + ( -6.15 + 104) 1 = 4.72 Nm 20.5 0.85 Lowering the load, motor torque when accelerating: M motor = - M b motor + K + ( - M b load + M load ) = -0.899 + ( -6.15 + 104) i 0.85 = 316 . Nm 20.5 (regenerative operation) Lowering the load, motor torque when decelerating: M motor = M v motor + K + ( M v load + M load ) = 0.899 + (615 . + 104) i 0.85 = 5.47 Nm 20.5 (regenerative operation) The highest motor torque is required when accelerating and hoisting the load, namely 98 % of the rated motor torque. The highest motor torque in regenerative operation is required when decelerating the drive while the load is being lowered. Calculating the maximum braking power The maximum motor output in regenerative operation occurs when the drive starts to decelerate from the maximum motor speed while the load is being lowered. Pbr motor max = M motor v nmotor max 9550 =- 5.47 1362 = -0.78 kW 9550 Selecting the drive converter The highest required motor torque approximately corresponds to the rated motor torque. An appropriate overload capability is required to start the hoisting drive, i. e. the drive converter should be able to provide approximately double the rated motor current as IU max. 52 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks Selected drive converter: 6SE7016-1EA61 PV n=2.2 kW; IV n= 6.1 A; IV max=8.3 A Closed-loop frequency control type Dimensioning the brake resistor The brake resistor is used when the load is lowered. Velocity-time diagram for hoisting and lowering v Hoisting No-load interval Lowering No-load interval v max t v2 v min tk t v1 - v t min ts t total -v ~ ~ ~ ~ tb max T Maximum distance travelled when hoisting and lowering: hmax = 0.8 m Cycle time for hoisting and lowering: T = t cycle = 90 s Siemens AG SIMOVERT MASTERDRIVES - Application Manual 53 3 Various special drive tasks 09.99 Time when the drive operates at the crawl velocity: t s = 0.4 s (specified) Accelerating time from 0 to vmax: v max 0.348 = = 0.6 s a max 0.58 tb = Decelerating time from vmax to vmin: tv 1 = v max - v min 0.348 - 0.025 = = 0.557 s a max 0.58 Deceleration time from vmin to 0: vmin 0.025 = = 0.043 s a max 0.58 tv 2 = From the area in the v-t diagram, hmax is given by: hmax = vmin t v 2 vmax tb v +v + vmax t k + max min t v 1 + vmin t s + 2 2 2 The unit travels at constant velocity vmax for: hmax - tk = 0.8 - = vmin t v 2 vmax tb vmax + vmin - t v 1 - vmin t s - 2 2 2 vmax 0.348 0.6 0.348 + 0.025 0.025 0.043 - 0.557 - 0.025 0.4 - 2 2 2 = 167 . s 0.348 Thus, the total travelling time for hmax is given by: t to ta l = tb + t k + t v 1 + t s + t v 2 = 0.6 + 167 . + 0.557 + 0.4 + 0.043 = 3.27 s 54 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks Max. braking power rating while accelerating from 0 to vmax: Pbr W max (0v max ) = Pbr motor max ( 0vmax ) motor = = M motor b nmotor max 9550 motor 316 . 1362 0.75 = 0.338 kW 9550 Braking power while travelling at constant velocity, vmax: Pbr W const ( v =vmax ) = Pbr motor const ( v =vmax ) motor = = M motor const nmotor max 9550 motor 4.31 1362 0.75 = 0.46 kW 9550 Max. braking power while decelerating from vmax to vmin: Pbr W max (vmax v min ) = Pbr motor max ( vmax vmin ) motor = = M motor v nmotor max motor 9550 5.47 1362 0.75 = 0.585 kW 9550 Min. braking power while decelerating from vmax to vmin: Pbr W min ( vmax vmin ) = Pbr motor min ( vmax vmin ) motor = = M motor v nmotor min 9550 motor 5.47 97.9 0.75 = 0.042 kW 9550 Braking power while travelling at constant velocity vmin: Pbr W const ( v =vmin ) = Pbr motor const ( v =vmin ) motor = = M motor const nmotor min 9550 motor 4.31 97.9 0.75 = 0.033 kW 9550 Max. braking power while decelerating from vmin to 0: Pbr W max (v min 0) = Pbr motor max ( vmin 0) motor = M motor v nmotor min 9550 motor = Pbr motor min ( vmax vmin ) motor = 0.042 kW Siemens AG SIMOVERT MASTERDRIVES - Application Manual 55 3 Various special drive tasks 09.99 Braking diagram P br W Hoisting Interval Lowering Interval 0.585 KW Constant velocity (v=vmax ) 0.460 kW Deceleration 0.338 kW Acceleration Constant vel. (v=vmin ) Deceleration ~ ~ ~ ~ 0.042 kW 0.033 kW 0.6 s t 1.67 s 0.557 s 0.4 s 0.043 s 3.27 s 90 s Braking energy for one cycle (corresponds to the area in the braking diagram) Precise calculation: Wbr = Pbr W max ( vmax vmin ) + Pbr W min ( vmax vmin ) 1 Pbr W max ( 0vmax ) t b + Pbr W const ( v =vmax ) t k + t v1 2 2 + Pbr W const ( v = vmin ) t s + = 1 Pbr W max ( vmin 0) t v 2 2 1 0.585 + 0.042 0.338 0.6 + 0.46 1.67 + 0.557 2 2 +0.033 0.4 + 1 0.042 0.043 = 1.058 kWs 2 In this particular case, the following estimation can be made: Wbr Pbr W const ( v =vmax ) (tb + t k ) + Pbr W const ( v =vmin ) t s 0.46 (0.6 + 1.67) + 0.033 0.4 = 1.057 kWs 56 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks The following must be valid for the brake resistor: Wbr 1057 . = = 0.012 kW Pbr cont . 90 T With Pbr cont . = P20 36 (with internal braking resistor) the following is obtained 36 0.012 = 0.432 kW P20 Thus, the smallest braking unit is selected with P20 = 5 kW (6SE7018-0ES87-2DA0). The internal brake resistor is adequate. Thermally checking the motor As the motor is only subject to a maximum load of approx. rated torque, and is only powered-up for approx. 7 s from a total of 90 s, a thermal analysis is not required. Positioning error as result of the brake application time tE After the brake motor is powered-down by inhibiting the drive converter pulses, there is first a brake application delay time tE before the mechanical brake becomes effective. The drive is shutdown at fU=0 and n0 (the slip speed is neglected). The load drops, accelerating due to its own weight, as there is no opposing motor torque. After the brake has been applied, standstill is reached after time tv mech. t v mech v ~ ~ tE t Motor powered-down - vmax E sv sE Brake applied Siemens AG SIMOVERT MASTERDRIVES - Application Manual 57 3 Various special drive tasks 09.99 The positioning error calculation is sub-divided into range 1 (acceleration due to the load itself) and range 2 (deceleration due to the mechanical brake). Calculating the acceleration caused by the load itself (range 1) For Mmotor=0: M motor = 0 = - M b motor + K + ( - M b load + M load ) i and 0 = - ( J motor + J K ) b motor - J load b motor + M load 2 i i The motor angular acceleration is then given by: b motor = M load i J motor + J K + J load 2 i = 104 0.85 20.5 0.85 0.0035 + 0.00028 + 0.53 20.52 = 889 s -2 and for the load acceleration: ab = b motor D 889 01 . = = 2.17 m / s 2 2 20.5 2 i The distance moved while the system accelerates due to the load: sE = 1 1 ab t E2 = 2.17 0.04 2 = 0.0017 m 2 2 Velocity after the brake application time has expired: vmax E = ab t E = 2.17 0.04 = 0.087 m / s Calculating the deceleration as result of the mechanical brake (range 2) For Mmotor=Mmech: M motor = M mech = M v motor + K + ( M v load + M load ) i and M mech = ( J motor + J K ) v motor + J load 58 v motor + M load 2 i i Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks The angular deceleration of the motor is then given by: M mech - M load v motor = J motor i + J K + J load 2 i = 14.5 - 104 0.85 20.5 0.85 0.0035 + 0.00028 + 0.53 20.52 = 2100 s -2 and the deceleration of the load itself: av = v motor D 2100 01 . . m / s2 = = 512 2 20.5 2 i Delay time as result of the mechanical brake: t v mech = v max E av = 0.087 = 0.017 s 512 . Distance moved during the delay caused by the mechanical brake: sv = vmax E t v mech 2 = 0.087 0.017 0.00074 m 2 Due to the delayed application of the mechanical brake after the brake motor has been powered down - time tE - the load moves a total distance of stotal = sE + sv = 0.0017 + 0.00074 = 0.00244 m = 2.44 mm downwards. This load-dependent positioning error must be taken into account by the positioning system. Siemens AG SIMOVERT MASTERDRIVES - Application Manual 59 3 Various special drive tasks 3.1.4 09.99 High-bay racking vehicle This application involves a high-bay racking vehicle with traversing drive, elevating drive and telescopic drive. In this case, only the traversing- and elevating drives are investigated. 1PA6 compact induction motors are to be used. Traversing drive Drive data Mass to be moved m = 5500 kg Load wheel diameter D = 0.34 m Specific traction resistance wF = 0.02 Gearbox ratio i = 16 Mech. efficiency = 0.75 Max. traversing velocity vmax = 2.66 m/s Max. acceleration amax = 0.44 m/s2 Max. traversing distance smax = 42.6 m No-load interval after a traversing sequence tP = 10 s Traversing at constant velocity Resistance force, load torque: FW = m g wF = 5500 9.81 0.02 = 1079.1 N M load = FW D 2 = 1079.1 60 0.34 = 183.45 Nm 2 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks Motor output, motor torque: Pmotor = FW vmax 1079.1 2.66 = = 383 . kW 103 0.75 10 3 M motor = M load 183.45 = = 15.29 Nm i 16 0.75 Motor speed at vmax: nmotor max = i nload max = i v max 60 2.66 60 = 16 = 2391 RPM D 0.34 Calculating the load torques when accelerating and decelerating Angular acceleration of the load wheel: load = a max 2 2 = 0.44 = 2.59 s -2 D 0.34 Load moment of inertia referred to the load wheel: D 2 ) 2 0.34 2 = 5500 ( ) = 158.95 kgm2 2 J load = m ( Load accelerating torque: M b load = J load load = 158.95 2.59 = 4114 . Nm As deceleration is the same as acceleration, the following is true: M v load = M b load Motor selection The following motor is selected (refer to Calculating the motor torques): 1PA6 103-4HG., Pn=7.5 kW, Mn=31 Nm, nn=2300 RPM, In=17 A, I=8.2 A Jmotor=0.017 kgm2, motor=0.866 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 61 3 Various special drive tasks 09.99 Calculating the motor torques when accelerating and decelerating Motor accelerating- and decelerating torques: = J motor load i M b motor = M v motor = 0.017 2.59 16 = 0.704 Nm Motor torque when accelerating: M motor = M b motor + ( M b load + M load ) = 0.704 + (411.38 + 183.45) 1 i 1 = 50.27 Nm 16 0.75 Motor torque when decelerating: M motor = - M v motor + ( - M v load + M load ) = -0.704 + ( -411.38 + 183.45) i 0.75 = -11.39 Nm 16 (regenerative operation) The highest motor torque is required when accelerating, namely 162 % of the rated motor torque. The moments of inertia of the gearbox, coupling, brake and other rotating masses have been neglected. Accelerating- and decelerating time: tb = t v = v max 2.66 = =6s a max 0.44 Investigating the condition so that the wheels do not spin when starting The following must be true so that the drive wheels don't spin when starting: F R > Fb Whereby: Fb = m ab max = 5500 0.44 = 2420 N 62 (accelerating force) Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks For two axes where one is driven, and when assuming that the driven axis carries half of the vehicle weight: F R = 1 1 m g = 5500 9.81 0.15 = 4047 N 2 2 (sum of the frictional forces) Thus, the above condition is fulfilled. Calculating the maximum braking power The maximum motor output in regenerative operation occurs when the drive starts to decelerate from the maximum motor speed. Pbr motor max = M motor v nmotor max 9550 =- 1139 . 2391 = -2.85 kW 9550 Selecting the drive converter For the highest required motor torque, the approximate motor current is given by: I motor max ( kn = nmot max nmot n M motor max M motor n = 2 2 2 2 )2 ( I motor n - I n ) kn + I n 2391 = 1,0396 2300 1 k n2 field weakening factor With I n = 8.2 A , the following is obtained: I motor max ( 50.27 2 1 2 ) (17 2 - 8.2 2 ) 10396 . + 8.2 2 2 = 26.3 A . 31 10396 for tb=6 s Selected drive converter: 6SE7022-6EC61 PV n=11 kW; IV n=25.5 A; IV max=34.8 A Closed-loop speed control Siemens AG SIMOVERT MASTERDRIVES - Application Manual 63 3 Various special drive tasks 09.99 Dimensioning the brake resistor The brake resistor is used during deceleration. As the braking characteristics are the same for both forwards- and reverse motion, only forwards motion has to be investigated. Velocity-time diagram for forwards motion v Forwards motion No-load interval vmax tb tv tk tp t t total vmax T t p = 10 s t v = tb = 6 s smax = 42.6 m Thus, the total time for the maximum travel (distance) is obtained as follows: t total = 2 tb + = 26+ smax - vmax tb vmax 42.6 - 2.66 6 = 22 s 2.66 Time for traversing at constant velocity: t k = t total - 2 t b = 22 - 2 6 = 10 s 64 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks Cycle time for forwards- and reverse travel: T = t total + t p = 22 + 10 = 32 s Maximum braking power for the brake resistor: Pbr W max = Pbr motor max motor Inv = 2.85 0.866 0,98 = 2.42 kW Braking diagram P br W Forwards motion No-load interval 2.42 kW Deceleration 6s 10 s t 22 s 32 s Braking energy for a cycle (corresponds to the area in the braking diagram): Wbr = 1 1 Pbr W max t v = 2.42 6 = 7.26 kWs 2 2 The following must be valid for the brake resistor: Wbr 7.26 = = 0.227 kW Pbr cont . T 32 With Pbrcont . = P20 36 (with an internal brake resistor) Siemens AG SIMOVERT MASTERDRIVES - Application Manual 65 3 Various special drive tasks 09.99 the following is obtained 36 0.227 = 817 . kW P20 Thus, a braking unit is selected with P20 = 10 kW (6SE7021-6ES87-2DA0). The internal brake resistor is adequate. Thermally checking the motor Torque characteristic for forwards motion Forward motion M No-load interval 50.27 Nm motor 15.29 Nm 6s 6s 10 s t 10 s -11.39 Nm 32 s The RMS torque is obtained from the torque characteristic as follows: M RMS = = 50.27 2 6 + 15.29 2 10 + 1139 . 2 6 6 + 10 + 6 + 10 18278.7 23 = 23.9 Nm 66 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks The calculated RMS torque is lower than the rated motor torque with 30 Nm. Thus, operation is thermally permissible. When calculating the RMS value, the slight field-weakening range is neglected. Elevating drive Drive data Mass to be moved m = 1350 kg Load wheel diameter D = 0.315 m Gearbox ratio i = 63.3 Mech. efficiency = 0.7 Max. elevating velocity vmax = 0.6 m/s Max. acceleration amax = 0.6 m/s2 Max. height hmax =6m No-load interval after elevating or lowering tP =8s Elevating the load at constant velocity Elevating force, load torque: FH = m g = 1350 9.81 = 132435 . N M load = FH D 2 = 13243.5 0.315 = 2085.9 Nm 2 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 67 3 Various special drive tasks 09.99 Motor output, motor torque: Pmotor = . 0.6 FH v max 132435 . kW = = 1135 3 10 0.7 103 M motor = M load 20859 . = = 471 . Nm i 633 . 0.7 Motor speed at vmax: nmotor max = i nload max = i v max 60 0.6 60 . = 633 = 2303 RPM D 0.315 Lowering the load at constant velocity Motor output, motor torque: Pmotor = - FH v max 13243.5 0.6 = - 0.7 = -5.56 kW 3 10 10 3 M motor = M load 2085.9 = 0.7 = 23.07 Nm 633 . i (regenerative operation) Calculating the load torques when accelerating and decelerating Angular acceleration of the load wheel: load = a max 2 2 = 0.6 = 3.81 s -2 D 0.315 Load moment of inertia referred to the load wheel: D 2 ) 2 0.315 2 = 1350 ( ) = 33.49 kgm2 2 J load = m ( 68 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks Load accelerating torque: M b load = J load load = 33.49 381 . = 127.58 Nm As deceleration is the same as acceleration, the following is true: M v load = M b load Motor selection The following motor is selected (refer to Calculating the motor torques): 1PA6 103-4HG., Pn=7.5 kW, Mn=31 Nm, nn=2300 RPM, In=17 A, I=8.2 A Jmotor=0.017 kgm2, motor=0.866 Calculating the motor torques when accelerating and decelerating Accelerating- and decelerating torque for the motor: M b motor = M v motor = J motor load i = 0.017 381 . 633 . = 41 . Nm Elevating the load, motor torque when accelerating: M motor = M b motor + ( M b load + M load ) = 41 . + (127.58 + 2085.9) 1 i 1 = 54.05 Nm 633 . 0.7 Elevating the load, motor torque when decelerating: M motor = - M v motor + ( - M v load + M load ) = -41 . + ( -127.58 + 2085.9) 1 i 1 = 401 . Nm 633 . 0.7 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 69 3 Various special drive tasks 09.99 Lowering the load, motor torque when accelerating: M motor = - M b motor + ( - M b load + M load ) = -41 . + ( -127.58 + 2085.9) i 0.7 = 17.55 Nm 633 . (regenerative operation) Lowering the load, motor torque when decelerating: M motor = M v motor + ( M v load + M load ) = 41 . + (127.58 + 2085.9) i 0.7 = 28.58 Nm 62.3 (regenerative operation) The highest motor torque is required when accelerating, while elevating the load, namely 174 % of the rated motor torque. The highest motor torque in regenerative operation is required when decelerating while the load is being lowered. The accelerating- or decelerating component in this case is low with respect to the load torque. The moments of inertia for gearbox, brake and coupling have been neglected. Accelerating and decelerating time: tb = t v = v max 0.6 = =1s a max 0.6 Calculating the maximum braking power The maximum motor output in regenerative operation occurs when the drive starts to decelerate from the maximum motor speed while the load is being lowered. Pbr motor max = M motor v nmotor max 9550 =- 28.58 2303 = -6.89 kW 9550 Drive converter selection For the highest motor torque, the approximate motor current is given by: I motor max ( 70 M motor max M motor n 2 2 2 ) 2 ( I motor n - I n ) + I n Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks With I n = 8.2 A , the following is obtained: I motor max ( 54.05 2 ) (17 2 - 8.2 2 ) + 8.2 2 = 27.2 A 31 for tb=1 s 5% is added for saturation effects, so that a max. motor current of 28.5 A is obtained. Selected drive converter: 6SE7022-6EC61 PV n=11 kW; IV n=25.5 A; IV max=34.8 A Closed-loop speed control Dimensioning the brake resistor The brake resistor is used when lowering the load. Velocity-time diagram for elevating and lowering v Elevating v No-load interval Lowering No-load interval max t tb t t b t k tv total t k tv tp t tp total - v max T tp = 8 s t v = tb = 1 s hmax = 6 m Siemens AG SIMOVERT MASTERDRIVES - Application Manual 71 3 Various special drive tasks 09.99 Thus, a total time is obtained for hmax: t total = 2 tb + = 2 1 + hmax - vmax tb vmax 6 - 0.6 1 = 11 s 0.6 Time for travel at constant velocity: t k = t total - 2 tb = 11 - 2 1 = 9 s Cycle time for elevating and lowering: T = 2 (t total + t p ) = 2 (11 + 8) = 38 s Max. braking power for the brake resistor when decelerating: Pbr W max v = Pbr motor max v motor Inv = 6.89 0.866 0.98 = 5.85 kW Max. braking power for the brake resistor when accelerating: Pbr W max b = Pbr motor max b motor Inv = = M motor b nmotor max 9550 motor Inv 17.55 2303 0.866 0.98 = 359 . kW 9550 Braking power for the brake resistor when travelling at constant velocity: Pbr W const = Pbr motor const motor Inv = 5.56 0.866 0.98 = 4.72 kW 72 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks Braking diagram P Elevating br W Interval Interval Lowering Constant velocity 5.85 kW 4.72 kW Deceleration 3.59 kW Acceleration 11 s 8s 1s 9s 1s 8s t 11 s 38 s Braking energy for a cycle (corresponds to the area in the brake diagram): Wbr = 1 1 Pbr W max b tb + Pbr W const t k + Pbr W max v t v 2 2 = 1 1 359 . 1 + 4.72 9 + 5.85 1 = 47.2 kWs 2 2 The following must be valid for the brake resistor: Wbr 47.2 = = 124 . kW Pbr cont . 38 T With Pbr cont . = P20 4.5 (for an external brake resistor) the following is obtained 4.5 124 . = 5.58 kW P20 Thus, a braking unit is selected with P 20 = 10 kW (6SE7021-6ES87-2DA0), with an external brake resistor (6SE7021-6ES87-2DC0). Siemens AG SIMOVERT MASTERDRIVES - Application Manual 73 3 Various special drive tasks 09.99 Thermally checking the motor Motor characteristic when hoisting and lowering M Elevating No load interval motor Lowering No load interval 54.05 Nm 47.1 Nm 40.1 Nm 28.58 Nm 17.55 Nm 1s 9s 1s 8s 1s 11 s 23.07 Nm 9s 1s 8s t 11 s 38 s The RMS torque is obtained from the torque characteristic: M RMS = = . 2 9 + 401 . 2 1 + 17.552 1 + 2307 . 2 9 + 28.58 2 1 54.052 1 + 471 2 (1 + 9 + 1 + 8) 30410 38 = 28.3 Nm The calculated RMS torque is less than the rated motor torque with 31 Nm. Thus, operation is thermally permissible. 74 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks Using a rectifier/regenerative feedback unit with inverters As an alternative, instead of using 3 drive converters with braking units, a multi-motor drive system can also be configured using a rectifier/regenerative feedback unit, DC bus and 3 inverters. Selecting the rectifier/regenerative feedback unit The low output of the telescopic drive with respect to the traversing- and elevating drive, can be neglected when dimensioning the rectifier/regenerative feedback unit, as the telescopic drive is always operational just by itself. On the other hand, the traversing- and elevating drive can operate simultaneously. When selecting the rectifier/regenerative feedback unit, it is assumed, that under worst case conditions, when accelerating, the maximum power and therefore the maximum DC link current of both drives occur simultaneously. Maximum motor power and maximum DC link current for the traversing drive: Pmotor max traverse = I DClink max traverse = M motor max traverse nmax traverse 9550 = 50.27 2391 = 12.59 kW 9550 12.59 103 = = 27.47 A 135 . 400 0.866 0.98 Pmotor max traverse U ZK motor traverse inv Max. motor power and maximum DC link current for the elevating drive: Pmotor max hoist = I DClink max hoist = M motor max hoist nmax hoist 9550 Pmotor max hoist U DClink motor hoist inv = 54.05 2303 = 1303 . kW 9550 13.03 103 = = 28.43 A 135 . 400 0.866 0.98 Thus, the maximum DC link current is given by: I DClink max total = I DClink max traverse + I DClink max hoist = 27.47 + 28.43 = 55.9 A Selected rectifier/regenerative feedback unit: 6SE7024-1EC85-1AA0 Pn=15 kW; IDC link n=41 A; IDC link max=56 A Siemens AG SIMOVERT MASTERDRIVES - Application Manual 75 3 Various special drive tasks 09.99 Selecting the regenerative feedback transformer When selecting the regenerative feedback transformer, the RMS DC link current for the rectifier/regenerative feedback unit is calculated in the regenerative mode. Previously, to dimension the braking resistors, the braking powers Pbr W =Pbr DC link in the DC link, were already calculated for the traversing- and elevating drive. In this case, the inverter efficiency was neglected. Thus, the calculation is on the safe side. Using the equation I DClink gen = Pbr DClink VDClink the RMS value in regenerative operation is obtained: I DClink gen RMS = Pbr DClink RMS VDClink The RMS value of the DC link braking power is obtained with the braking powers Pbr W , specified in the brake diagrams for the traversing- and elevating drive: Pbr DClink RMS = 1 1 1 2.42 2 6 + 359 . 2 1 + 4.72 2 9 + 5.852 1 3 3 3 = 2.45 kW 38 To calculate the RMS value of the non-constant segments, the following equation is used. i +1 P 2 br W i dt = 1 ( Pbr2 W i + Pbr2 W i +1 + Pbr W i Pbr W i+1 ) t i 3 The cycle time for the elevating drive, 38 s, was used as time interval. The following is now obtained for the RMS value of the DC link current: I DClink gen RMS = 2.45 103 = 4.54 A 135 . 400 The permissible RMS value is as follows for a regenerative feedback transformer with 25% duty ratio: I DClink RMS permissible = I DClink n 0.92 25 = I DClink n 0.46 = 41 0.46 = 18.86 A 100 Thus, a regenerative feedback transformer 4AP2795-0UA01-8A with a 25% duty ratio is sufficient. In addition, a 4% uk line reactor 4EP3900-5US is required. 76 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3.1.5 3 Various special drive tasks Hoisting drive for a 20 t gantry crane This application involves a four-line hoisting unit with 2 cable runs. The test load is 25 t. Previously, a 6-pole wound-rotor motor (US: slipring motor) was used, directly connected to the line supply with Pn=21.5 kW; this has now been replaced by a 6-pole induction motor with Pn=22 kW. Cable drum i, G IM D Load Hoisting unit data Cable drum diameter D = 0.6 m Gearbox ratio i = 177 Gearbox efficiency G = 0.9 Permissible load Q = 20 t Test load Qmax = 25 t Accelerating time, decelerating time tb, tv = 10 s Rated motor output Pn = 22 kW Rated motor torque Mn = 215 Nm Rated motor speed nn = 975 RPM Motor moment of inertia JMotor = 0.33 kgm2 Maximum hoisting height hmax =5m Siemens AG SIMOVERT MASTERDRIVES - Application Manual 77 3 Various special drive tasks 09.99 Hoisting the load at constant velocity Hoisting force and load torque at the cable drum: FH = mload g 20 10 3 9.81 = = 98100 N 2 2 M load = FH D 0.6 = 98100 = 29430 Nm 2 2 Circumferential speed at the cable drum: vdrum max = = n motor D 2 nn motor D = i 2 i 60 2 2 975 0.6 = 0.173 m / s 177 60 2 Motor output, motor torque: Pmotor = FH vdrum max M motor = G 10 3 = 98100 0173 . = 18.86 kW 0.9 10 3 M load 29430 = = 184.7 Nm i G 177 0.9 For a 25 t test load, a motor output of 23.57 kW and a motor torque of 231 Nm are obtained. The selected 22 kW motor is sufficient, as it involves a drive with intermittent duty. Lowering the load at constant velocity Motor output, motor torque: Pmotor = - M motor = 78 FH vdrum max 10 3 G = - 98100 0173 . 0.9 = -15.27 kW 3 10 (regenerative operation) M load 29430 G = 0.9 = 149.6 Nm i 177 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks Calculating the motor torques when accelerating Motor angular acceleration: motor = = n motor 2 nn motor = tb 60 tb 2 975 = 10.2 s -2 60 10 Accelerating torque for motor+gearbox: M b motor + gearbox = ( J motor + J gearbox ) motor = (0.33 + 0) 10.2 = 3.37 Nm (Jgearbox neglected) Moment of inertia of the load referred to the cable drum: J load = = 1 D mload ( ) 2 4 2 1 0.6 20 103 ( ) 2 = 450 kgm2 4 2 Moment of inertia of the cable drum: JTr=20 kgm2 (estimated) Accelerating torque for the load+wire drum: M b load + drum = ( J load + J drum ) drum = ( J load + J drum ) = (450 + 20) motor i 10.2 = 27.08 Nm 177 Hoisting the load, motor torque when accelerating: M motor = M b motor + gearbox + ( M b load + drum + M load ) = 3.37 + (27.08 + 29430) 1 i G 1 177 0.9 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 79 3 Various special drive tasks 09.99 Due to the relatively long accelerating time, the accelerating torques are low with respect to the load torque. Selecting the motor and drive converter Selected motor: 1LA5 207-6 Pn=22 kW; Imotor n=42.5 A; Mn=215 Nm; motor=90.8%; Jmotor=0.33 kgm2 Selected drive converter: 6SE7027-2ED61 PV n=37 kW; IV n=72 A; IV max=98 A Closed-loop frequency control The maximum current of the 37 kW drive converter is 98 A. Thus, the 22 kW motor (Imotor n= 42.5 A) can also easily start even with a suspended test load of 25 t. The starting current must be set high enough, so that the load doesn't slip when the brake is released and there are no problems at the transition from open-loop controlled operation to closed-loop controlled operation. It is practical to set the slip frequency for the brake release somewhat higher than the rated slip frequency. Rated motor slip frequency: f slip n = n S - nn 1000 - 975 50 Hz = 50 = 1.25 Hz nS 1000 As a result of the relatively long accelerating time from 10 s to 50 Hz, the brake enable signal can be simply realized using the signal fV fx, with fx = 2 Hz. In the still available time from 0.4 s to when the brake is released, the flux will have had time to establish itself. Dimensioning the brake resistor The brake resistor is used when the load is lowered. As result of the low influence of the accelerating- and decelerating torques, only the braking power due to the constant load torque when the load is being lowered is taken into account. The worst case situation is investigated, when the nominal load is lowered from the full height. 80 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks Velocity-time diagram when lowering the load from hmax=5 m v load t total tb tv t - v load max vload max = vdrum max 2 = 0.0865 m / s t v = tb = 10 s Total time for the load to be lowered: t total = t br = 2 tb + = 2 10 + hmax - vload max tb vload max 5 - 0.0865 10 = 67.8 s 0.0865 Max. braking power of the braking resistor: Pbr W max = = FH vTr max 103 G motor 98100 0.173 0.9 0.908 = 13.87 kW 103 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 81 3 Various special drive tasks 09.99 Braking diagram P br W 13.87 kW 10 s 10 s t 67.8 s Braking energy for a 90 s cycle (corresponds to the area in the braking diagram): Wbr = Pbr W max (t total - t b ) = 1387 . (67.8 - 10) = 8017 . kWs The following must be true for the brake resistor: Wbr 801.7 = = 8.91 kW Pbr cont . 90 T With Pbr cont . = P20 4.5 (for an external brake resistor) the following is obtained 8.91 4.5 = 40.1 kW P20 Thus, a braking unit is selected with P 20 = 50 kW (6SE7028-0EA87-2DA0) with an external brake resistor (6SE7028-0ES87-2DC0). 82 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3.1.6 3 Various special drive tasks Elevator (lift) drive This drive involves an elevator with a 1:1 suspension system, and a load capability of 1080 kg for 14 people. The elevator moves through a total height of 23 m with 8 floors. The elevator approaches the floors directly without crawl using a closed-loop position control function. Drive pulley i, G D IM Counter-weight Cable Suspended cable Cabin + load Elevator data Driving pulley diameter D = 0.64 m Gearbox ratio i = 35 Gearbox efficiency G = 0.69 Load capability Q = 1080 kg Max. weight of the elevator cabin Fmax = 1200 kg Max. counter-weight (Fmax+0.5Q) Gmax = 1740 kg 1/2 suspended cable weight 1/2 mHK = 28.75 kg Cable weight mcable = 48.3 kg Operational velocity vmax = 1.25 m/s Acceleration amax = 0.9 m/s2 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 83 3 Various special drive tasks 09.99 Rated motor output Pn = 13.5 kW Rated motor current IMotor n = 33.5 A Rated motor torque Mn = 94 Nm Rated motor speed nn = 1455 RPM Motor efficiency Motor = 0.89 Motor moment of inertia JMotor = 0.21 kgm2 Moment of inertia of the hoist (referred to the drive pulley) JW = 105.7 kgm2 Moment of inertia of the handwheel (referred to the drive pulley)JH = 27 kgm2 Elevator height range = 23 m FH Ascending with a full load and at constant velocity n Tr M load Hoisting force: FH = (1.02 (mF + mQ ) - 0.97 mG + mcable ) g D FH = (102 . (1200 + 1080) - 0.97 1740 + 48.3) 9.81 = 6730.6 N Cable (1,02; 0,97: additional factors) Load torque, motor torque: M load = FH M motor D 0.64 = 6730.6 = 2153.8 Nm 2 2 G v F+Q M 2153.8 = load = = 89.18 Nm i G 35 0.69 Motor output: Pmotor = FH v max 6730.6 1.25 = = 12.2 kW G 103 0.69 103 The existing 13.5 kW motor is sufficient, as the drive operation is intermittent (refer to the thermal check later in the text). Motor speed at vmax: nmotor max = i ndrum max = i 84 v max 60 . 60 125 = 35 = 1305 RPM D 0.64 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks Descending with full load and constant velocity n Tr D Motor torque, motor output: M motor Pmotor M load M 2153.8 = load G = 0.69 = 42.46 Nm i 35 FH Cable F v 6730.6 1.25 = - H 3max G = - 0.69 = -5.8 kW 10 10 3 F+Q v (regenerative operation) G Calculating the motor torques when accelerating and decelerating Motor angular acceleration: motor = i a max 2 2 = 35 0.9 = 98.44 s -2 D 0.64 Accelerating torque for the motor: M b motor = J motor motor = 0.21 98.44 = 20.67 Nm Load moment of inertia referred to the drive pulley: m linear = 1.02 (mF + mQ + mG + mSeil + 1 mHK ) 2 = 1.02 (1200 + 1080 + 1740 + 48.3 + 28.75) = 4180 kg D J load = mlinear ( ) 2 2 = 4180 ( 0.64 2 ) = 428 kgm2 2 Accelerating torque for load+hoist+handwheel: M b load +W + H = ( J load + JW + J H ) motor i = (428 + 105.7 + 27) 98.44 = 1577 Nm 35 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 85 3 Various special drive tasks 09.99 As deceleration is the same as acceleration, the following is valid: = M b motor M v motor M v load +W + H = M b load +W + H Ascending with full load, motor torque when accelerating: M motor = M b motor + ( M b load +W + H + M load ) = 20.67 + (1577 + 2153.8) 1 i G 1 = 175.2 Nm 35 0.69 Ascending with full load, motor torque when decelerating: M motor = - M v motor + ( - M v load +W + H + M load ) = -20.67 + ( -1577 + 21538 . ) 1 i G 1 = 3.2 Nm 35 0.69 Descending with full load, motor torque when accelerating: M motor = - M b motor + ( - M b load +W + H + M load ) = -20.67 + ( -1577 + 2153.8) G i 0.69 = -9.3 Nm 35 Descending with full load, motor torque when decelerating: M motor = M v motor + ( M v load +W + H + M load ) = 20.67 + (1577 + 2153.8) G i 0.69 = 94.22 Nm 35 (regenerative operation) The highest motor torque is required when accelerating while ascending with a full load. The highest motor torque in the regenerative mode is when decelerating while descending with a full load. Accelerating- and decelerating times: tb = t v = 86 v max 125 . = = 1.39 s a max 0.9 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks Calculating the maximum braking power The maximum motor output in regenerative operation occurs when the drive starts to decelerate from the maximum motor speed while descending with full load. Pbr motor max = M motor v nmotor max 9550 =- 94.22 1305 = -12.88 kW 9550 Selecting the drive converter Selected drive converter: 6SE7026-0ED61 PV n=30 kW; IV n=59 A; IV max=80.5 A T300 technology board with closed-loop elevator control The drive converter is adequately dimensioned, as the rated converter current corresponds to 176 % of the rated motor current. For the highest required motor torque, the approx. motor current is given by: I motor max ( M motor max M motor n 2 2 2 ) 2 ( I motor n - I n ) + I n With I n = 0.35 I motor n (assumption), the following is obtained: I motor max ( 175.2 2 ) (335 . 2 - 0.352 335 . 2 ) + 0.352 335 . 2 = 59.7 A 94 Dimensioning the brake resistor When dimensioning the brake resistor, the worst case situation is investigated which is when the elevator is descending with a full load through all the floors. An 8 s no-load interval is assumed between ending a descent and starting to make an ascent. Siemens AG SIMOVERT MASTERDRIVES - Application Manual 87 3 Various special drive tasks 09.99 Velocity-time diagram for a descent and an ascent through all of the floors (max. height) v Descent v No-load interval Ascent No-load interval max t t total tk b tv tp tb t k tv tp t t total - v max T tp = 8 s . s t v = t b = 139 hmax = 23 m Thus, the following total time is obtained for travelling through the full height: t total = 2 t b + hmax - v max t b 23 - 125 . 139 . = 2 139 . + = 19.79 s v max 125 . Constant travel velocity time: t k = t total - 2 t b = 19.79 - 2 139 . = 17 s Cycle time for the descent and ascent: T = 2 (t total + t p ) = 2 (19.79 + 8) = 5558 . s Max. braking power of the brake resistor: Pbr W max = Pbr motor max motor = 12.88 0.89 = 1146 . kW Brake resistor power when the elevator is moving at a constant velocity: Pbr W const = Pbr motor const mot = 58 . 0.89 = 516 . kW 88 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks Braking diagram P br W Descent No-load interval 11.46 kW Ascent No-load interval 5.16 kW 8s 19.8 s 8s t 17 s 1.39 s 1.39 s 19.8 s 55.58 s Braking energy for a cycle (corresponds to the area in the braking diagram): Wbr 1 = Pbr W const t k + Pbr W max t v 2 1 = 516 . 17 + 1146 . 139 . 2 = 87.7 + 7.96 = 95.7 kWs The following must be valid for the brake resistor: Wbr 95.7 = = 1.72 kW Pbr cont . 5558 . T With Pbr cont . = P20 4.5 (with an external brake resistor) the following is obtained 4.5 172 . = 7.74 kW P20 Thus, a braking unit is selected with P 20 = 10 kW (6SE7021-6ES87-2DA0) with an external brake resistor (6SE7021-6ES87-2DC0). Siemens AG SIMOVERT MASTERDRIVES - Application Manual 89 3 Various special drive tasks 09.99 Thermally checking the motor This elevator makes 180 ascents/descents per hour, i. e. 20 s per ascent/descent including 8 s noload interval. Thus, when thermally checking the motor, a duty cycle at full load through 5 floors has been assumed. Height difference between the bottom and top floors (5 floors): h5 = 5 23 = 14.4 m 8 Total time to travel from the first to the fifth floor (through 5 floors): t total = 2 tb + h5 - vmax tb 14.4 - 125 . 139 . = 2 139 = 12.91 s . + 125 . vmax Cycle time for a descent and an ascent: T = 2 (t total + t p ) = 2 (12.91 + 8) = 4182 . Time for which the elevator moves with continuous velocity: t k = t total - 2 tb = 12.91 - 2 139 . = 1013 . s Torque characteristics when descending and ascending Ascending M motor No-load interval Descending No-load interval 175.2 Nm 94.22 Nm 89.18 Nm 42.46 Nm 1.39 s 3.2 Nm 8s 10.13 s 1.39 s 1.39 s 10.13 s -9.3 Nm 8s t 1.39 s 41.82 s 90 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks The RMS torque is obtained from the torque characteristic: . + 89.18 2 1013 . + 3.2 2 139 . + 9.32 139 . + 42.46 2 1013 . + 94.22 2 139 . 175.2 2 139 = 2 (12.91 + 0.33 8) M RMS = 153967.7 311 . = 70.4 Nm Calculating the average speed: nmot average = nmot i A + nmot i E 2 te + k f t P ti 1 nmax t b 2 + nmax t k = 2 2 tb + tk + k f t P 1 1305 139 . 2 + 1305 1013 . 2 = = 967 RPM 2 139 . + 1013 . + 0.33 8 The average motor speed is greater than half the rated motor speed (control range 1:2) and the calculated RMS torque is less than the motor rated torque with 94 Nm. Thus, operation is thermally permissible. Siemens AG SIMOVERT MASTERDRIVES - Application Manual 91 3 Various special drive tasks 3.1.7 09.99 Traction drive along an incline In this case, it involves a rack and pinion drive. The gradient is 3 degrees. At standstill, the vehicle is held at standstill by the motor holding brake. Motor Gearbox Pinion Rack Drive data Mass to be moved (fully laden) m = 3500 kg Mass to be moved (empty) m = 3000 kg Pinion diameter D = 0.12 m Angle of inclination (gradient) = 3 degrees Specific traction resistance wF = 0.05 Mech. efficiency = 0.9 Max. velocity vmax = 0.2 m/s Max. acceleration ab max = 0.8 m/s2 Max. deceleration av max = 0.8 m/s2 Max. distance moved smax = 2.4 m Cycle time tcycle = 210 s 92 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks Specified motion characteristic v Ascending vmax 1 2 1 t p5 t p4 t p6 ~ ~ t p3 t p2 t t p1 3 _v 5 4 6 max Descending T Distances travelled (as specified): Range 1: s1= 1.4 m with m=3500 kg (ascending) Range 2: s2= 1 m with m=3000 kg (ascending) Total 2.4 m Range 3: s3= 0.15 m with m=3000 kg (descending) Range 4: s4= 0.65 m with m=3000 kg (descending) Range 5: s5= 0.2 m with m=3000 kg (descending) Range 6: s6= 1.4 m with m=3500 kg (descending) Total 2.4 m No-load intervals (as specified): tp1=tp5=2 s tp2=tp3=tp4=8 s Siemens AG SIMOVERT MASTERDRIVES - Application Manual 93 3 Various special drive tasks 09.99 Accelerating time, decelerating time: tb = t v = vmax 0.2 = = 0.25 s ab max 0.8 With t ki = si - 0.5 v max (t b + t v ) vmax the following times are obtained when moving at a constant velocity: t k1 = 6.75 s t k 2 = 4.75 s t k 3 = 0.5 s tk4 = 3 s t k 5 = 0.75 s t k 6 = 6.75 s Thus, the time for the complete distance moved is given by: i =5 i =6 i =1 i =1 t ascent +descent = t pi + t ki + 6 (tb + t v ) = 28 + 22.5 + 3 = 53.5 s With the cycle time (specified) T = t cycle = 210 s the remaining no-load interval is given by: t p 6 = T - t ascent + descent = 210 - 535 . = 156.5 s Precise positioning is only possible using a servo drive due to the short accelerating- and decelerating times. 94 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks Ascending at constant velocity (loaded) v F Fload H F F N W m g Component of gravity down the incline, normal component of force, drag: FH = m g sin = 3500 9.81 sin 30 = 1797 N FN = m g cos = 3500 9.81 cos 30 = 34288 N FW = FN wF = 34288 0.05 = 1714.4 N Force due to the load, load torque: Fload , ascend = FH + FW = 1797 + 1714.4 = 35114 . N M load , ascend = Fload = 3511.4 D 2 0.12 = 210.7 Nm 2 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 95 3 Various special drive tasks 09.99 Descending at constant velocity (loaded) v F F load H F F N W m g Force due to the load, load torque: Fload descend = FH - FW = 1797 - 1714.4 = 82.6 N M load descend = Fload = 82.6 D 2 0.12 = 4.95 Nm 2 Calculating the load torques when accelerating and decelerating (loaded) Angular acceleration of the pinion: b pinion = ab max 2 2 = 0.8 = 13.33 s -2 D 0.12 Load moment of inertia: D 0.12 2 J load = m ( ) 2 = 3500 ( ) = 12.6 kgm2 2 2 As acceleration is the same as deceleration, the following is valid for the load torques: M b load = M v load = J load b pinion = 12.6 13.33 = 168 Nm 96 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks Selecting the motor The following motor is empirically selected (refer to calculating the motor torques): 1FT6 041-4AF7 with gearbox i=81, nn=3000 RPM, Mn(100)=2.2 Nm, In(100)=1.7 A, motor=0.86 Jmotor+brake=0.00039 kgm2, Jgearbox=0.0008 kgm2 Motor speed at vmax: nmotor max = i v max 60 0.2 60 = 81 = 2578.3 RPM D 0.12 Calculating the motor torques while the vehicle is moving at constant velocity (loaded) Ascending: M motor = M load ascend 1 1 = 210.7 = 2.89 Nm i 81 0.9 Descending: M motor = M load descend 0.9 = 4.95 = 0.055 Nm i 81 (regenerative operation)1) 1) Regenerative operation is obtained from the various signs for torque and speed Calculating the motor torques when accelerating and decelerating (loaded) Motor accelerating- and decelerating torques: M b motor = M v motor = J motor i b pinion = 0.00039 81 13.33 = 0.42 Nm The accelerating- and decelerating torques for the gearbox, referred to the pinion: M b gearbox = M v gearbox = J gearbox i 2 b pinion = 0.0008 812 13.33 = 70 Nm Siemens AG SIMOVERT MASTERDRIVES - Application Manual 97 3 Various special drive tasks 09.99 Ascending, motor torque when accelerating: M motor = M b motor + ( M b load + M b gearbox + M load ascend ) = 0.42 + (168 + 70 + 210.7) 1 i 1 = 6.58 Nm 81 0.9 Ascending, motor torque when decelerating: M motor = - M v motor + ( - M v load - M v gearbox + M load ascend ) 1) = -0.42 + ( -168 - 70 + 210.7) 1 i sign (...) 1 = -0.72 Nm 81 0.9 -1 (regenerative operation) Descending, motor torque when accelerating: M motor = - M b motor + ( - M b load - M b gearbox + M load descend ) 1) = -0.42 + ( -168 - 70 + 4.95) sign (...) i 0.9 -1 = -3.62 Nm 81 Descending, motor torque when decelerating: M motor = M v motor + ( M v load + M v gearbox + M load descend ) = 0.42 + (168 + 70 + 4.95) i 0.9 = 312 . Nm 81 (regenerative operation) 1) For a negative expression in brackets, changes to -1 The highest motor torque is required when accelerating as the vehicle ascends. The highest motor torque in regenerative operation is required when decelerating as the vehicle descends. The selected motor is adequately dimensioned, as it can be overloaded up to approx. 7 Nm at nmax=2578.3 RPM and 400 V supply voltage. The steady-state holding brake torque corresponds to the motor torque when descending at constant velocity: M hold = 0.055 Nm 98 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks Selecting the drive converter For the highest required motor torque, the motor current is given by: I motor max = I motor n M motor max M motor n = 17 . 6.58 = 5.08 A 2.2 (hardly any saturation effect for this motor) Selected drive converter: 6SE7013-0EP50 (Compakt Plus type) PV n=1.1 kW; IV n=3 A, IV max=9 A (300% overload capability) Dimensioning the brake resistor The brake resistor is used when the vehicle decelerates as it ascends and when it descends at constant velocity and when decelerating. In the following calculation, the slight decrease in the motor torque during motion without load is neglected (thus the calculation is on the safe side). Max. braking power for the brake resistor when decelerating from v max to 0: Pbr W max (vmax 0) = Pbr motor max ( vmax 0) motor = = M motor v ( vmax 0) nmotor max 9550 motor 0.72 2578.3 0.86 = 0.168 kW 9550 Braking power for the brake resistor while moving at constant velocity with -vmax: Pbr W const ( v =- vmax ) = Pbr motor const ( v =- vmax ) motor = = M motor const ( v =- vmax ) nmotor max 9550 motor 0.055 2578.3 0.86 = 0.013 kW 9550 Max. braking power for the brake resistor when decelerating from -vmax to 0: Pbr W max (-vmax 0) = Pbr motor max ( - vmax 0) motor = = M motor v ( - vmax 0) nmotor max 9550 motor 312 . 2578.3 0.86 = 0.724 kW 9550 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 99 3 Various special drive tasks 09.99 Brake diagram P br W Ascending (2x) Descending (4x) 0.724 kW ~ ~ 0.013 kW ~ ~ 0.168 kW t k3,4,5,6 tv t tv T Braking energy for a cycle (corresponds to the area in the braking diagram): 1 1 Wbr = 2 Pbr W max ( vmax 0) t v + Pbr W const ( v =- vmax ) (t k 3 + t k 4 + t k 5 + t k 6 ) + 4 Pbr W max ( - vmax0 ) t v 2 2 1 1 = 2 0.168 0.25 + 0.013 (0.5 + 3 + 0.75 + 6.75) + 4 0.724 0.25 = 0.545 kWs 2 2 The following must be valid for the brake resistor: Wbr 0.545 = = 0.0061 kW Pbr cont . 90 T As the cycle time with 121 s is greater than 90 s, T is set to 90 s. With Pbr cont. = P20 4,5 (for Compact Plus, only an external brake resistor) the following is obtained 4.5 0.0061 = 0.027 kW P20 Thus, the smallest braking resistor is selected with P 20 = 5 kW (6SE7018-0ES87-2DC0). 100 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks Thermally checking the motor In the following calculation, the slight decrease in the motor torque when the vehicle is not carrying a load is neglected (thus, the calculation is on the safe side). Torque characteristics Ascending (2x) Descending (4x) M motor 6.58 Nm 3.12 Nm 2.89 Nm 0.055 Nm ~ ~ t k1,2 tb tb v ~ ~ t tv -0.72 Nm t t k3,4,5,6 -3.62 Nm T The RMS torque is obtained from the torque characteristic as follows t k 1 + t k 2 = 6.75 + 4.75 = 115 . s t k 3 + t k 4 + t k 5 + t k 6 = 0.5 + 3 + 0.75 + 6.75 = 11 s Thus, M RMS = 2 (6.58 2 + 0.72 2 ) 0.25 + 2.89 2 115 . + 4 (3.62 2 + 312 . 2 ) 0.25 + 0.0552 11 = 0.82 Nm 6 0.25 2 + 115 . + 11 + 1 (2 + 8 + 8 + 8 + 2 + 156.5) Siemens AG SIMOVERT MASTERDRIVES - Application Manual 101 3 Various special drive tasks 09.99 For the 1FT6 motor, factor kf is set to 1 for the no-load intervals. The calculated RMS torque is less than the rated motor torque with 2.2 Nm. Thus, operation is thermally permissible. Information regarding the calculation Due to the tedious calculations required when trying-out various motors and gearbox ratios, it is recommended that a spreadsheet program such as Excel is used. 102 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks 3.2 Winder drives 3.2.1 General information Reel i, Motor F Z =const. G D core D D max n mot, Mmot V master=const. nW Mode of operation of an axial winder A material web, with width b, is wound or unwound at constant velocity and tension. v 60 D reel speed [RPM] v web velocity [m/s] D actual reel diameter [m] nW = nW min = v 60 Dmax minimum speed of the reel at D=Dmax [RPM] nW max = v 60 DKern max. speed of the reel at D=Dcore [RPM] FZ = z b web tension [N] z web tension/m [N/m] b web width [m] Siemens AG SIMOVERT MASTERDRIVES - Application Manual 103 3 Various special drive tasks MW = FZ D v 60 1 = FZ 2 2 nW MW max = FZ Dmax 2 M W min = FZ PW = Dcore 2 FZ v 103 09.99 winding torque (proportional to 1/nW ) [Nm] max. winding torque at D=Dmax [Nm] min. winding torque at D=Dcore [Nm] winding power (constant) [kW] Winding Unwinding M W , PW D max M W max D core M W ~ 1/nW PW = const. MW min n W min n W max n W Winding torque and winding power at v = const. and z = const. 104 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks Influence of the friction Further, a speed-dependent frictional torque is present due to the bearing friction and other effects. Thus, the winding torque is increased when winding and reduced when unwinding. MR speed-dependent frictional torque [Nm] MW res = MW + M R resulting winding torque when winding [Nm] MW res = MW - M R resulting winding torque when unwinding [Nm] Motor torque, motor output (steady-state operation) With i= nmotor Dcore nmotor max = nW v 60 G gearbox ratio gearbox efficiency the following is obtained for the motor: MW + M R i G M motor = Pmotor = MW + M R nW 9550 G M motor = - Pmotor = - MW - M R G i MW - M R nW G 9550 motor torque when winding [Nm] motor output when winding [kW] (energy flow from the motor to the load) motor torque when unwinding [Nm] motor output when unwinding [kW] (energy flow from the load to the motor) Siemens AG SIMOVERT MASTERDRIVES - Application Manual 105 3 Various special drive tasks 09.99 Acceleration, deceleration When accelerating and decelerating the reel at constant web tension, additional accelerating- and deceleration torques are present. The effective moment of inertia consists of a fixed component (motor, gearbox/coupling = supplementary, winding core) and a variable component (reel as hollow cylinder). J F = J motor + J sup plementary + JV = J core i2 b W 4 103 ( D 4 - Dcore ) 32 fixed moment of inertia referred to the motor [kgm2] shaft variable reel moment of inertia [kgm2] (as a function of reel diameter D) and JV = 1 2 mW ( D 2 + Dcore ) 8 JV max = [kgm2] b W 4 4 10 3 ( Dmax - Dcore ) maximum reel moment of inertia 32 [kgm2] 1 2 2 mW ( Dmax + Dcore ) 8 [kgm2] and JV max = JV min = 0 for D = Dcore min. reel moment of inertia b, D in m; in kg/dm3; m in kg Accelerating- and decelerating torque for the motor + gearbox/coupling referred to the motor shaft: M b , v motor +sup plement = ( J motor + J sup plement ) i 2 v D tb, v [Nm] Accelerating- and decelerating torque for core+reel referred to the reel: M b , v core +reel = ( J core + JV ) 106 2 v D tb, v [Nm] Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks tb accelerating time from 0 to v [s] tv decelerating time from v to 0 [s] When accelerating and decelerating, the variable moment of inertia is assumed to be constant. For constant tension during acceleration and deceleration, the motor torque is given by: Winding, motor torque when accelerating from 0 to v: M motor = M b motor + sup plement + ( M b core + reel + M W + M R ) 1 i G [Nm] Winding, motor torque when decelerating from v to 0: M motor = - M v motor + sup plement + ( - M v core+ reel + M W + M R ) 1) 1 i G [Nm] 1) If the expression in the brackets is < 0, the factor 1/ must be changed to (the G G deceleration component is predominant) Unwinding, motor torque when accelerating from 0 to v: M motor = M b motor + sup plement + ( M b core +reel - M W + M R ) 2 ) G i [Nm] Unwinding, motor torque when decelerating from v to 0: M motor = - M v motor + sup plement + ( - M v core+ reel - M W + M R ) G i [Nm] 2) If the expression in brackets is > 0, factor must be changed to 1/ (acceleration G G component is predominant) Siemens AG SIMOVERT MASTERDRIVES - Application Manual 107 3 Various special drive tasks 09.99 The motor output is obtained from the motor torque as follows Pmotor = M motor nmotor 9550 [kW] Motor selection The motor is selected according to the maximum torque at nmin and the maximum speed. For additional accelerating- and decelerating torques, it may be possible to utilize the overload capability of the drive. As the winding torque is proportional to 1/n, then it is possible to use the field-weakening range, with Mperm. 1/n. Further, there must be sufficient safety margin to the stall torque in the field-weakening range. The winding power PW is constant over the speed. Thus, the motor must provide this output at nmin. M motor M stall 1,3 M perm. M motor stat. Field weakening range n min nn n max n motor Motor torque under steady-state operating conditions and permissible torque for S1 duty 108 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks Closed-loop winder control There are several ways of providing constant tension: * indirect closed-loop tension control (without tension transducer) In this case, the constant tension is implemented using a torque reference value input adapted by the diameter. It is important to accurately compensate the frictional- and accelerating torques, as these parameters cannot be compensated without using a tension transducer. * direct closed-loop tension control with dancer roll The tension is entered using a position-controlled dancer roll. * direct closed-loop tension control with tension transducer The tension actual value is sensed using a tension transducer and is appropriately controlled. Note: An Excel program is available to dimension winders: "McWin - Motorcalculation for Winder". Also refer to ASI Information E20125-J3001-J409-X-7400 from September 94. Siemens AG SIMOVERT MASTERDRIVES - Application Manual 109 3 Various special drive tasks 3.2.2 09.99 Unwind stand with closed-loop tension control using a tension transducer This application involves an unwind stand with "flying" roll change. Before the reel is changed, the new reel is prepared for splicing. When the old reel falls below a specific diameter, the new reel is accelerated until the circumferential speed corresponds to the web velocity of the paper. The reel support mechanism then rotates into the splice position. The new reel is automatically spliced, and the old reel retracted. Unwind stand data Max. reel diameter Dmax = 1.27 m Core diameter Dcore = 0.11 m Web width b = 1.7 m Material density W = 0.93 kg/dm3 Max. web velocity vmax = 15 m/s Web tension z = 100 N/m Acceleration tb = 40 s Stop (coast down) tv = 40 s Friction MR = 20 Nm at Dmax Gearbox ratio i = 1 to 3 Steady-state operation nW min = v max 60 15 60 = = 226 RPM Dmax 127 . nW max = v max 60 15 60 = = 2604 RPM Dcore 0.11 FZ = z b = 100 1.7 = 170 N M W max = FZ Dmax 127 . = 170 = 108 Nm 2 2 M W min = FZ Dcore 0.11 = 170 = 9.4 Nm 2 2 110 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 PW = 3 Various special drive tasks FZ v max 170 15 = = 2.55 kW 10 3 10 3 Gearbox ratio i = 2 is selected The gearbox efficiency G is assumed to be 1 (belt-driven gearbox, pinion wheel) nmotor min = i nW min = 2 226 = 452 RPM nmotor max = i nW max = 2 2604 = 5208 RPM M motor max = - M motor min = - MW max - M R i MW min - M R i =- =- 108 - 20 = -44 Nm 2 9.4 - 0 = -4.7 Nm 2 The frictional torque at nW min was assumed to be 0. The motor torques are negative, as this involves an unwind stand (regenerative operation). Calculating the maximum motor torques when accelerating and decelerating b W 4 4 10 3 ( Dmax - Dcore ) 32 1.7 0.93 3 = 10 (1.27 4 - 0.114 ) = 403.76 kgm 2 32 JV max = J core 0 (cardboard core) JF = J motor + J sup plement 0.08 kgm2 (estimated) Relationships at D=Dmax Mb motor +sup plement = ( J motor + J sup plement ) i = 0.08 2 M b core +reel 2 v max Dmax t b 2 15 = 0.094 Nm 127 . 40 = ( J core + J roll ) = (0 + 403.76) 2 v max Dmax tb 2 15 = 238.4 Nm 1.27 40 Mv = M b (as tb = tv) Siemens AG SIMOVERT MASTERDRIVES - Application Manual 111 3 Various special drive tasks 09.99 Accelerating from 0 to vmax (without tension, with friction): M motor = M b motor + sup plement + ( M b core + reel - M W max + M R ) = 0.094 + ( 238.4 - 0 + 20) 1 i 1 2 = 0.094 + 129.2 = 129.3 Nm Decelerating from Vmax to 0 (with tension, with friction): M motor = - M v motor + sup plement + ( - M v core + reel - M W max + M R ) = -0.094 + ( -238.4 - 108 + 20) 1 i 1 2 = -0.094 - 163.2 = -1633 . Nm The highest motor torque is required when decelerating with tension applied. At Dmax the accelerating- or decelerating torque for the motor+supplement is essentially of no significance. Relationships at D=Dcore: M b motor +sup plement = ( J motor + J sup plement ) i = 0.08 2 M b core +reel 2 v max Dcore t b 2 15 = 109 . Nm 011 . 40 0 ( JV = 0, J core 0) Mv = M b (as tb = tv) Accelerating from 0 to vmax (without tension, without friction): M motor = M b motor + sup plement + ( M b core+ reel - M W min + M R ) = 109 . + ( 0 - 0 + 0) 112 1 i 1 = 109 . Nm 2 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks Decelerating from vmax to 0 (with tension, without friction): = - M v motor + sup plement + ( - M v core + reel - M W min + M R ) M motor = -109 . + ( 0 - 9.4 + 0) 1 i 1 = -5.79 Nm 2 The highest motor torque is required when decelerating with tension. Motor selection The motor selected as result of the calculated maximum motor torques: 1PA6 133-4HD. Pn=13.5 kW; Mn=112 Nm; nn=1150 RPM; n1=2500 RPM; In=29 A; I=13 A; Jmotor=0.076 kgm2 When decelerating from vmax to 0 (with tension, with friction), at nmin, the motor must provide 1633 . = 146 . x 112 rated motor torque for 40 s. The maximum motor output is then given by: Pmotor = M motor nmotor min 9550 = 163.3 452 = 7.73 kW 9550 The 44 Nm steady-state winding torque is permissible for the motor at nmin. To check the motor in the field-weakening range, the permissible motor output at nmax, with 130 % safety margin, is given by: Pperm. nmax = Pn n1 2500 = 135 . = 6.48 kW nmax 5208 Permissible motor torque at nmax and 130 % safety margin from the stall torque: M perm. nmax = Pperm. nmax 9550 nmax = 6.48 9550 = 119 . Nm 5208 Thus, the motor can also provide the required torque of 5.79 Nm, at nmax, when decelerating with tension. The maximum motor output is then given by: Pmotor = M motor nmotor max 9550 = 5.79 5208 = 316 . kW 9550 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 113 3 Various special drive tasks 09.99 Selecting the inverter and the rectifier/regenerative feedback unit The maximum motor current in dynamic operation is given by: M motor max dyn I motor max dyn ( M motor n 2 2 2 ) 2 ( I motor n - I n ) + I n With I n = 13 A , the following is obtained: I motor max dyn ( 1633 . 2 ) ( 29 2 - 132 ) + 132 = 40 A 112 The maximum motor current for winding operation is: I motor max stat ( I motor max stat ( M motor max stat M motor n 2 2 2 )2 ( I Motor n - I n ) + I n 44 2 ) ( 29 2 - 132 ) + 132 = 16,5 A 112 Thus, an inverter is selected (2x): 6SE7023-4TC20 PV n=15 kW; IV n=34 A; IV max=46.4 A T300 technology board with closed-loop winder control Infeed/regenerative feedback unit: 6SE7024-1EC85-1AA0 Pn=15 kW; IDC link n=41 A Autotransformer with 25% duty ratio: 4AP2795-0UA01-8A (25% is sufficient, as the power when winding is only 2.55 kW) 4% uk line reactor: 4EP3900-5UK 114 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks 0 0 1000 2000 3000 4000 5000 6000 -20 Torque in Nm -40 Limited by M stall M mot stat M S1 charac. -60 -80 P = const -100 M mot = Mn -120 Speed in RPM Diagram for the winding torque (M mot stat) and the MS1 characteristic Siemens AG SIMOVERT MASTERDRIVES - Application Manual 115 3 Various special drive tasks 09.99 M mot = 2 Mn 250 200 150 Limited by Mstall M mot b M mot v+tens. M max perm. 100 Torque in Nm 50 0 0 1000 2000 3000 4000 5000 6000 -50 -100 -150 -200 -250 Speed in RPM Diagram for the accelerating torque without tension (M mot b), the deceleration torque with tension (M mot v+tension) and the Mperm max characteristic 116 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks 3.2.3 Winder with closed-loop tension control using a tension transducer This involves a winder with accumulator. The material web, running at max. 50 m/min, is woundup, tension controlled. Just before the final diameter is reached, the reel support assembly is rotated, the winder is stopped, the paper web cut, and the second reel spliced to the empty core. The material accumulator must accept the web coming from the machine during this time (approx. 15-20 s). After the empty reel starts, this runs, speed-controlled, 17.8 m/min faster than the web speed of the machine. Thus, the material accumulator is emptied in approx. 60 s. After this, the closed-loop tension controlled mode is selected. Winder data Max. reel diameter Dmax =1m Core diameter Dcore = 0.12 m Max. reel weight mW = 870 kg Max. web velocity in operation vmax B = 50 m/min = 0.833 m/s Max. web velocity when retrieving paper vmax L = 67.8 m/min = 1.13 m/s from the accumulator Web tension FZ = 600 N Acceleration tb =2s Fast stop tv =2s Material thickness d = 2 mm Steady-state operation nW min = nW max = v max B 60 Dmax v max L 60 Dcore = 0.833 60 = 15.9 RPM 1 = 113 . 60 = 180 RPM 012 . Dmax 1 = 600 = 300 Nm 2 2 D 0.12 = FZ core = 600 = 36 Nm 2 2 MW max = FZ M W min Siemens AG SIMOVERT MASTERDRIVES - Application Manual 117 3 Various special drive tasks PW = FZ v max B 10 3 09.99 600 0.833 = 0.5 kW 10 3 = Selected gearbox ratio, i = 20 The gearbox efficiency G is assumed to be 0.9 nmotor min = i nW min = 20 15.9 = 318 RPM nmotor max = i nW max = 20 180 = 3600 RPM M motor max = MW max M motor min = MW min i G i G = 300 = 16.7 Nm 20 0.9 = 36 = 2 Nm 20 0.9 Calculating the maximum motor torques when accelerating and decelerating JV max = 1 2 2 mW ( Dmax + Dcore ) 8 = 1 870 (12 + 0.12 2 ) = 110.3 kgm 2 8 J core 0 (cardboard core) JF = J motor + J sup plement 0.03 kgm 2 (estimated) Relationships at D=Dmax M b motor +sup plement = ( J motor + J sup plement ) i 2 v max B Dmax tb 2 0.833 = 0.03 20 = 0.5 Nm 1 2 M b core +reel = ( J core + J reel ) 2 v max B Dmax tb 2 0.833 = (0 + 110.3) = 91.92 Nm 1 2 Mv = M b (as tb = tv) 118 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks Accelerating from 0 to vmax B (with tension): M motor = M b motor + sup plement + ( M b core+ reel + M W max ) = 0.5 + (91.92 + 300) 1 i G 1 20 0.9 = 0.5 + 2177 . = 22.27 Nm Decelerating from vmax B to 0 (with tension): M motor = - M v motor + sup plement + ( - M v core + reel + M W max ) = -0.5 + ( -91.92 + 300) 1 i G 1 20 0.9 = -0.5 + 1156 . = 1106 . Nm Decelerating from vmax B to 0 (without tension, e. g. material web broken): M motor = -0.5 + ( -91.92) 0.9 20 (the factor 1/G is changed into G, as the expression in brackets is < 0) = -0.5 - 4.14 = -4.64 Nm The highest motor torque is required when the winder is accelerating with tension. The highest regenerative motor torque is required when decelerating without tension. The accelerating- and decelerating torque for motor+supplement practically plays no role at Dmax. Relationships at D=Dcore: Mb motor +sup plement = ( J motor + J sup plement ) i = 0.03 20 M b core +reel 2 v max L Dcore tb 2 113 . = 5.65 Nm 0.12 2 0 ( JV = 0, J core 0) Mv = M b (as tb = tv) Siemens AG SIMOVERT MASTERDRIVES - Application Manual 119 3 Various special drive tasks 09.99 Accelerating from 0 to vmax L (with tension): M motor = M b motor + sup plement + ( M b core+ reel + M W min ) = 5.65 + (0 + 36) 1 i G 1 = 7.65 Nm 20 0.9 Decelerating from vmax L to 0 (with tension): M motor = - M v motor + sup plement + ( - M v core+ reel + M W min ) = -5.65 + (0 + 36) 1 i G 1 20 0.9 = -5.65 + 2 = -3.65 Nm Decelerating from vmax L to 0 (without tension): M motor = -5.65 Nm The highest motor torque is required when accelerating with tension. The highest regenerative motor torque is required when decelerating without tension. Calculating the maximum braking power The following is valid for the motor output: Pmotor = Mmotor nmotor 9550 The maximum motor output in the regenerative mode occurs when decelerating without tension at maximum speed (i. e. at D=Dcore). Pbr motor max = 120 -5.65 3600 = -2.13 kW 9550 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks Motor selection The motor was selected as a result of the calculated maximum motor torques: 1PQ5 113-4AA20-Z Pn=4 kW; Mn=27 Nm; nn=1435 RPM; In=9.2 A; Mstall=3 Mn; motor=0.83; Jmotor=0.011 kgm2 Checking the motor in the field-weakening range at nmax Motor stall torque in the field-weakening range: M stall = M stall n ( nn 2 ) n Permissible motor torque at nmax and 130 % safety margin from the stall torque: M perm. = M stall M stall n n 3 27 1435 2 = ( n )2 = ( ) = 9.9 Nm 1.3 1.3 nmax 13 . 3600 The motor can also provide the demanded torque of 7.65 Nm when accelerating with tension, at nmax. Selecting the drive converter Selected drive converter: 6SE7021-0EA61 PV n=4 kW; In=10.2 A T300 technology board with closed-loop winder control Dimensioning the brake resistor The max. braking power for the brake resistor is: Pbr W max = Pbr motor max motor = 2.13 0.83 = 177 . kW Siemens AG SIMOVERT MASTERDRIVES - Application Manual 121 3 Various special drive tasks 09.99 Braking energy: Wbr = Pbr W max t v 2 = 1.77 2 . kWs = 177 2 Winding time: tW = 2 2 Dmax - Dcore . 2 12 - 012 = = 7.74 min 4 d v max 4 0.002 50 > 90 s Thus, the cycle time for braking is set to 90 s. The following must be valid for the brake resistor: Wbr 177 . = = 0.02 kW Pbr cont . 90 T With Pbr cont . = P20 36 (with an internal brake resistor) the following is obtained 36 0.02 = 0.72 kW P20 The lowest braking unit is selected with P 20 = 5 kW (6SE7018-0ES87-2DA0). The internal brake resistor is sufficient. 122 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3.2.4 3 Various special drive tasks Unwinder with 1FT6 motor A 1FT6086-8SF71-1AA0 motor has been selected for a winder with flying roll change. The dimensioning and correct motor selection are to be checked. The unwinder inverter is to be integrated into a multi-motor group. Unwind stand data Max.reel diameter Dmax = 0.8 m Core diameter Dcore = 0.09 m Reel weight m = 100 kg Max.web velocity vmax = 200 m/min Steady-state torque referred to the motor speed at Dmax Msteady-st. max = 31 Nm Required acceleration time tb = 0.5 s Required deceleration time tv = 0.5 s Gearbox ratio (belt driven) i =4 Moment of inertia of the mandrel J mandrel = 0.08 kgm2 Diameter of the large gear wheel on the mandrel (112 teeth) DZ mandrel = 290 mm Diameter of the small gear wheel at the motor (28 teeth) DZ mot = 72.5 mm Width of the gear wheels b = 38 mm Rated output Pn = 9.7 kW Rated speed nmot n = 3000 RPM Moment of inertia Jmot = 0.00665 kgm2 Efficiency mot = 0.91 Torque constant kT0 = 1.33 Nm/A Stall torque M0 = 35 Nm Stall current I0 = 26.4 A Max. permissible torque Mmax = 90 Nm Max. permissible current Imax = 84 A Gear wheel material: Steel Motor data Siemens AG SIMOVERT MASTERDRIVES - Application Manual 123 3 Various special drive tasks 09.99 Max. web velocity in m/s: v max = 200 = 333 . m/ s 60 Web tension: FZ = M stat max i 2 2 = 31 4 = 310 N Dmax 0.8 Steady-state operation PW = - FZ v max 310 333 . . kW =- = -103 1000 1000 nmot min = i v max 60 . 60 333 = 4 = 318.3 RPM Dmax 0.8 nmot max = i v max 60 333 . 60 = 2829.4 RPM = 4 Dcore 0.09 Steady-state motor torque as a function of the motor speed: M mot steady - state = - M steady - state max nmot min n mot Determining the motor torques when accelerating and decelerating as a function of the motor speed Max. moment of inertia of the full roll: J v max = D 100 0.8 2 0.09 2 m Dmax 2 (( ) + ( core ) 2 ) = (( ) + ( ) ) = 81 . kgm2 2 2 2 2 2 2 Estimating the additional moments of inertia: J Z mandrel 1 1 0.29 4 b 7.85 r 4 10 3 = 0.038 7.85 ( ) 10 3 = 0.066kgm 2 2 2 2 1 1 0.08 4 J Z mot b 7.85 r 4 103 = 0.038 7.85 ( ) 10 3 = 0.00026 kgm2 2 2 2 124 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks Thus, the supplementary moment of inertia, referred to the motor is given by: J sup plement = J mandrel + J Z mandrel i + J Z mot = 2 0.066 + 0.08 + 0.00026 0.01 kgm 2 2 4 Accelerating from 0 to vmax without tension (with tension) M mot b = ( J mot + J sup plement ) i + J v max 2 v max n mot Dmax t b nmot min 2 v max Dmax t b ( - M steady - state max 4 Dmax ( nmot min n mot 4 max D nmot min n mot component, motor+supplement 4 ) 4 - Dcore -D 4 core 1 i ) component, winder (component tension) The maximum motor torques when accelerating are obtained without tension. As this unwind stand has a flying roll change, the accelerating curve is calculated without tension. At a roll change, the full or partially full roll must first be accelerated in preparation for the flying roll change (where the two material webs are glued together). In operation, after standstill, acceleration only with tension. Decelerating from vmax to 0 with tension M mot v = -( J mot + J sup plement ) i 2 v max - J v max Dmax t v - M steady - state max 2 v max n mot Dmax t v nmot min 4 Dmax ( component, motor+supplement nmot min 4 ) 4 - Dcore nmot 1 4 4 Dmax - Dcore i n mot min n mot component, winder component tension Checking the selected motor The steady-state and dynamic torque characteristics, together with the motor limiting characteristics are plotted to check the selected motor. As can be seen from the following characteristics, operation with the selected motor is permissible. Siemens AG SIMOVERT MASTERDRIVES - Application Manual 125 3 Various special drive tasks 09.99 0 500 1000 1500 2000 2500 3000 -5 Torque in Nm -10 -15 Mmot steady-state MS1 charac. -20 -25 -30 -35 Motor speed in RPM Steady-state torque characteristic and Mperm S1 characteristic 100 80 60 Torque in Nm 40 20 0 0 500 1000 1500 2000 2500 3000 M mot b M mot v+tens. M max per. -20 -40 -60 -80 -100 Motor speed in RPM Acceleration characteristic without tension, deceleration characteristic with tension and Mperm max characteristic 126 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks Selecting the inverter The inverter is selected according to the peak current in dynamic operation and according to the highest motor current when winding. The highest motor current when winding must be less than the inverter base load current so that the overload capability of the inverter can be used. Thus, the following must be valid: I mot max dyn 1.6 I INV n I mot steady - state max 0.91 I INV n The motor current is given by, taking into account saturation: I mot = I mot = M mot kT0 b1 for M mot M 0 M mot M mot - M 0 2 M I ) (1 - max 0 )) kT0 b1 (1 - ( M max - M 0 M 0 I max for M mot > M 0 with b1 = 1 - b ( nmot 1,5 ) 6000 (b=0.1 for frame size <100, otherwise 0.15) The max. motor current when winding is 23.34 A. The max. motor current in dynamic operation is 52.77 A. Thus, the following inverter is selected: 6SE7023-4TC51 SIMOVERT MASTERDRIVES Motion Control PINV n=15 kW IINV n=34 A IINV max=54.5 A Siemens AG SIMOVERT MASTERDRIVES - Application Manual 127 3 Various special drive tasks 09.99 35 Motor current stat. in A 30 25 20 I mot steady-sta. I WR base load 15 10 5 0 0 500 1000 1500 2000 2500 3000 Motor speed in RPM Motor current when winding, inverter base load current 60 Motor current dyn. in A 50 40 I mot b I mot v+tens. I WR max 30 20 10 0 0 500 1000 1500 2000 2500 3000 Motor speed in RPM Motor currents when accelerating without tension and when decelerating with tension, maximum inverter current 128 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks DC link currents In order to integrate the inverter into a multi-motor drive group, the maximum DC link currents must be calculated in dynamic operation as well as the constant DC link current when winding. The DC link power is obtained from: PDC link = M mot 2 n mot 1 60 (mot INV )VZ with VZ = sign( M mot ) Thus, the DC link current is: I DC link = PDC link 1.35 U sup ly The maximum DC link powers, and therefore also the maximum DC link currents occur at the maximum speed. As acceleration without tension at maximum speed does not occur (empty roll), acceleration with tension is used when dimensioning (accelerating after a standstill). The following values are obtained with Usupply=400 V: I DC link b max = 3.92 A acceleration with tension I DC link v max = -6.53A deceleration with tension I DC link steady - state = -1.71A winding operation Siemens AG SIMOVERT MASTERDRIVES - Application Manual 129 3 Various special drive tasks 09.99 6 4 DC link current in A 2 0 0 500 1000 1500 2000 2500 3000 I DC link v+tens. I DC link b+tens. -2 -4 -6 -8 Motor speed in RPM DC link current when accelerating with tension and when decelerating with tension 130 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3.2.5 3 Various special drive tasks Unwind stand with intermittent operation An unwind stand is to be operated using two mechanically-coupled 1PA6 motors. The inverters for the two motors are supplied from a rectifier/regenerative feedback unit. The web velocity is not constant, but it is specified by the following characteristic. v vmax tb t tv tk T Unwind stand data Max. roll diameter Dmax = 1,25 m Core diameter Dcore = 0,125 m Roll weight m = 2000 kg Max. web velocity vmax = 600 m/min Web tension Fz = 500 N Required accelerating time tb =7s Required braking time tv =7s Constant-velocity time tk =1s Cycle time T = 25 s Gearbox ratio i =3 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 131 3 Various special drive tasks 09.99 Max. web velocity in m/s: v max = 600 = 10 m / s 60 Each motor should accept 50 % of the load. Thus, for the calculation, the weight of the roll and the web tension is always divided by two. Steady-state operation FZ 500 v max 10 PW = - 2 =- 2 = -2.5 kW 1000 1000 n mot min = i v max 60 10 60 = 3 = 458.4 RPM Dmax 1.25 n mot max = i v max 60 10 60 = 3 = 4583.7 RPM Dcore 0.125 Steady-state motor torque as a function of the motor speed: M mot stat = - M mot stat max nmot min nmot with M mot stat max = - Fz Dmax 1 500 1.25 1 =- = -52.08 Nm 2 2 i 2 2 3 Determining the motor torques when accelerating and decelerating as a function of the motor speed Max. moment of inertia of the full roll: J v max m 2000 D D 1.25 2 0.125 2 = 2 (( max ) 2 + ( core ) 2 ) = 2 (( ) +( ) ) = 197.27 kgm 2 2 2 2 2 2 2 Estimating the additional moments of inertia (referred to the motor): J sup plementary 0.1 kgm 2 132 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks Accelerating from 0 to vmax with tension M motb = ( J mot + J sup plementary ) i + J vmax 2 vmax Dmax tb - M stat max 2 vmax nmot Dmax tb nmot min 4 Dmax ( nmot min n mot 4 max 4 ) 4 - Dcore -D D 4 core component, motor+supplementary 1 i nmot min winding component tension component n mot Deceleration from vmax to 0 with tension M mot v = -( J mot + J sup pl . ) i - J vmax 2 vmax nmot Dmax tv nmot min 2 vmax Dmax tv - M stat max 4 Dmax ( nmot min n mot 4 max D 4 ) 4 - Dcore -D 4 core nmot min nmot component, motor+supplementary 1 i winding component tension component Selecting the motor The following motor is selected: 1PA6 137-4HF. Pn=25 kW; Mn=136 Nm; nn=1750 RPM; Mstall=485 Nm; In=56 A; I=23 A; Jmotor=0.109 kgm2 motor=0.902 When selecting the motor, it must be taken into account that the motor accelerating torque and the motor decelerating torque at each speed must be below the dynamic limiting curve of the motor. In addition, the RMS current, obtained from the traversing characteristic must be less or equal to the rated motor current at each speed. The highest torques, and therefore also the highest RMS current is obtained at nmot=nmot min as follows: M mot b max = 99.65 Nm max. accelerating torque M mot v max = -203.81 Nm max. decelerating torque Siemens AG SIMOVERT MASTERDRIVES - Application Manual 133 3 Various special drive tasks 09.99 With I mot ( M mot 2 2 2 2 ) ( I mot n - I n ) + I n M mot n (in the constant flux range) the maximum motor currents when accelerating, when decelerating and when unwinding with constant velocity are given by: I mot b max = 43.9 A I mot v max = 79.9 A I mot stat max = 30.2 A I mot v max I mot I mot b max I n I mot stat max tb tv tk t tp T Thus, the following RMS value is obtained: I RMS = 2 2 2 2 I mot b max tb + I mot stat max t k + I mot v max tv + I n t p T = 50.75 A This value lies below the rated motor current with 56 A. 134 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks To check the dynamic relationships, the dynamic limiting curve of the motor together with the accelerating and decelerating torque will be investigated. 2Mn 300 200 Limited by M stall M mot b+tens. M mot v+tens. M perm. max Torque in Nm 100 0 0 500 1000 1500 2000 2500 3000 3500 4000 4500 5000 -100 -200 -300 Motor speed in RPM Selecting the inverter The inverter is selected according to the peak current in dynamic operation and according to the highest RMS current. The following must therefore be valid: I mot max dyn 1.36 I INV n or I mot max dyn 1.6 I INV n (for 160 % overload capability) I mot rms max I INV n The max. RMS motor current is 50.75 A. The max. motor current in dynamic operation is 79.9 A. Thus, the following inverters are selected: 6SE7026-0TD61 SIMOVERT MASTERDRIVES Vector Control PINV n=30 kW; IINV n=59 A; IINV max=94.4 A (160 % overload capability) Siemens AG SIMOVERT MASTERDRIVES - Application Manual 135 3 Various special drive tasks 09.99 The 160 % overload capability can be used here, as the motor torque and therefore also the motor current decrease as the speeds increase, and as the maximum motor current is required in the constant flux range. Thus, the fact that field weakening is entered somewhat earlier (at 90 % rated speed) is of no importance. Selecting the rectifier/regenerative feedback unit The rectifier/regenerative feedback unit is dimensioned according to the maximum DC link current in dynamic operation and according to the highest RMS DC link current. The DC link currents for each inverter is calculated using the DC link power. PDClink = M mot 2 nmot 1 60 ( mot INV )VZ with VZ = sign( M mot ) Thus, the DC link current is given by: I DClink = PDClink 1.35 Vsup ply 15 10 DC link current in A 5 0 0 500 1000 1500 2000 2500 3000 3500 4000 4500 5000 I DC link b+tens. I DC link v+tens. -5 -10 -15 -20 Motor speed in RPM 136 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks The maximum DC link powers when accelerating and when decelerating, and therefore also the maximum DC link currents, occur at the minimum speed. Thus, with Vsupply=400 V, the following values are obtained for one motor: I DClink b max = 10.02 A max. DC link current when accelerating I DClink v max = -16.01 A max. DC link current when decelerating I DClink stat = -4.09 A DC link current at constant velocity I DC link b max motoring tk tv t tb I DC link stat regenerating I DC link v max T The following is valid for the RMS value when motoring, for one motor: 2 I DClink b max tb I DClink RMS mot = 3 T = 3.06 A The following is valid for the RMS value when generating for one motor: I DClink RMS gen = 2 I DClink stat tk + 2 I DClink v max tv 3 T Siemens AG SIMOVERT MASTERDRIVES - Application Manual = 4.96 A 137 3 Various special drive tasks 09.99 The rectifier/regenerative feedback unit is selected according to the following criteria for the summed currents of both motors: I DClink b max I I I = 2 10.02 = 20.04 1.36 I DClink R / Re n maximum value, motoring = 2 16.01 = 32.02 1.36 0.92 I DClink R / Re n maximum value, generating DClink v max DClink rms mot = 2 3.06 = 6.12 I DClink R / Re n RMS value, motoring DClink rms gen = 2 4.96 = 9.92 0.92 I DClink R / Re n RMS value, generating This results in a 15 kW rectifier/regenerative feedback unit with I DC link R/Re n=41 A. However, as the output of the rectifier/regenerative feedback unit should be at least 30% of the power rating of the connected inverter, a 37 kW rectifier/regenerative feedback unit is selected. 6SE7028-6EC85-1AA0 Pn=37 kW; IDC link n=86 A The permissible RMS value for a regenerative feedback transformer with 25% duty cycle is given by: I DClink rms perm. = I DClink R / Re n 0.92 25 = I DClink R / Re n 0.46 = 86 0.46 A = 39.56 A 100 As this value is greater than the regenerative RMS value, regenerative feedback transformer with a 25% duty cycle is sufficient. In addition, a 4% uk line reactor 4EU2451-4UA00 is required. 138 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks 3.3 Positioning drives 3.3.1 General information Positioning procedure Positioning is possible in two ways: * open-loop controlled * closed-loop controlled Open-loop controlled positioning With open-loop controlled positioning, during positioning, there is no feedback regarding the actual position. Thus, it is not guaranteed that the actual end position coincides with the desired end position. A reference point transmitter with a time-dependent velocity signal or a position sensor (e.g. Bero, opto-barrier) is required for open-loop controlled positioning. Effects on the accuracy * changing deadtimes (reading-in, processing) in the setpoint processing and when processing the control commands * reference point transmitter and position sensor resolution * play in the mechanical system (gearbox, joints etc.) * dependency on the load condition (e. g. motor slip) Siemens AG SIMOVERT MASTERDRIVES - Application Manual 139 3 Various special drive tasks 09.99 Example showing the influence of changing deadtimes when reading-in/processing control commands t max v vmax ~ ~ s max t Shutdown command (e.g. via BERO) The maximum response time to the shutdown command is, in the most unfavorable case, tmax. Thus, the maximum positioning error is: smax = vmax t max Positioning errors can be reduced by approaching the end position at crawl speed (rapid traverse, crawl speed positioning). v vmax t max smax ~ ~ vmin t Crawl speed starts 140 Shutdown command Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks The dwell time at vmin must be at least tmax, otherwise, positioning could be completed prematurely. The maximum positioning error is now only: smax = vmin t max The disadvantage when using a crawl speed is the longer positioning time. Closed-loop controlled positioning For closed-loop controlled positioning, the current position is compared to the reference value from the reference value transmitter. Deviations are equalized through the closed-loop position controller. sact sset xd Position controller vset Speed controller Actuator Motor Ref. point generator Block diagram for closed-loop controlled positioning xd: Position difference The reference point generator and closed-loop position controller can be externally set-up in the PLC or can also be integrated into a technology board in the drive converter. Instead of entering a motion characteristic via the reference point transmitter, the complete position difference can be fed to the closed-loop position controller. In this case, quantities vmax and amax during positioning are kept constant elsewhere (ramp-function generator, limiter, square-root function). Siemens AG SIMOVERT MASTERDRIVES - Application Manual 141 3 Various special drive tasks 09.99 The accuracy is influenced by the following factors * position encoder resolution * play in the mechanical system * deadtimes in the reference value processing * drive speed control range For high requirements regarding fast and precise positioning without overshoot, the drive must have the highest possible limiting frequency (> 40 Hz) and a high speed control range (> 1:1000). For applications, for example, feed drives, packing machines etc., servo drives are preferably used (SIMOVERT MC). If the requirements regarding dynamic performance are not as high, e. g. for high-bay racking vehicles, elevators etc., even standard AC drives can be used (SIMOVERT VC). The somewhat restricted speed control range can be compensated for using an appropriate gearbox ratio and a longer positioning time. Example for the influence of the encoder resolution when using a double-pulse encoder i Motor D/2 Encoder Gearbox s Block diagram Distance s moved during one motor revolution is: s= D i [mm] D drive wheel diameter i gearbox ratio 142 [mm] Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks Thus, smallest detectable position difference xdi at an edge change is given by: xdi = D i 4z z [mm] pulses per encoder revolution The lowest detectable position difference xdi should be at least smaller than the required accuracy by a factor of 4. If the drive is still to respond to a specific position difference xd , then it must still be possible to adjust a minimum velocity of vmin = xd k v kv [mm/s] position controller gain factor [s-1] The position controller gain must not be too high for stability reasons. It is dependent on the drive configuration (moment of inertia, gearbox play, deadtimes etc.) and the required characteristics. Values which can be implemented are: k v max = 10 to 20 for standard AC drives (SIMOVERT VC) k v max = 50 to 100 for servo drives (SIMOVERT MC) With the minimum velocity, the minimum motor speed is given by: nmin = i 60 vmin D [RPM] The required speed control range can be determined via the minimum motor speed. Siemens AG SIMOVERT MASTERDRIVES - Application Manual 143 3 Various special drive tasks 09.99 Fast positioning In order to achieve fast positioning, the drive must accelerate and brake at the maximum possible values, and the constant velocity range must be traversed at the maximum velocity. Example for fast positioning with tb=tv v s total v max tb tk tv t t total a a max t - a max With tb = t v = vmax amax (acceleration=deceleration) [s] amax max. acceleration [m/s2] vmax max. velocity [m/s] results in the distance travelled stotal: stotal = v max t total - 144 2 v max a max [m] Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks or for positioning time ttotal for the specified distance travelled: t total = v max stotal + a max v max [s] If the positioning time ttotal and the distance travelled stotal are specified, then vmax is given by: v max = a max t total a t - ( max total ) 2 - a max stotal 2 2 [m/s] The constant velocity time tk is obtained from: t k = t total - 2 v max a max [s] For the special case with tk=0, the trapezoidal velocity characteristic changes into a triangular characteristic. Example for fastest positioning with tb=tv, special case, tk=0 v v max s total tv tb t t total a amax t - a max Siemens AG SIMOVERT MASTERDRIVES - Application Manual 145 3 Various special drive tasks 09.99 With tb = t v = vmax amax (acceleration=deceleration) [s] the following is obtained: stotal 2 v max t total v max = = a max 2 [m] t total = 2 stotal 2 v max = v max a max [s] v max = a max t total = a max stotal 2 [m/s] Other acceleration characterisics In addition to fast positioning with constant acceleration, other acceleration characteristics are possible, especially for servo drives, e. g.: * Sawtooth acceleration This characteristic results in positioning with minimum motor losses. However, for the same positioning time, the maximum acceleration increases with respect to fastest positioning and the end position is rapidly approached (mechanical stressing). * Sinusoidal acceleration In this case, acceleration steps, which occur for fastest positioning or for sawtooth acceleration characteristics, are eliminated as well as the associated high mechanical stresses. However, disadvantages are the higher maximum acceleration for the same positioning time with respect to fastest positioning and the complex implementation. * Jerk-free acceleration characteristic The acceleration can be smoothly entered (jerk-free) to keep the stressing on the mechanical system low and to prevent oscillations being excited, i. e., no steps and transitions in the accelerating characteristic. However, disadvantages are the higher maximum acceleration for the same positioning time with respect to fastest positioning and the somewhat complex implementation. 146 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks 3.3.2 Traction drive with open-loop controlled positioning using Beros The maximum positioning error due to the changing deadtime when reading-in/processing control commands at the SIMOVERT VC drive converter are to be determined. Drive data Max. traversing velocity vmax = 36 m/min Min. traversing velocity (crawl velocity) vmin = 2 m/min Max. deadtime when processing the control commands tmax = 4xT0 = 4.8 ms Velocity-time diagram v v max ~ ~ v min t Bero 1 Bero 2 Bero 1: Changeover from vmax to vmin via fixed reference values (terminal strip) Bero 2: Off1 command via the terminal strip (the brake is energized at fU= 0) The maximum positioning error as result of the changing deadtime is: smax = vmin t max = 2 m / min 4.8 ms = 0.016 cm Siemens AG SIMOVERT MASTERDRIVES - Application Manual 147 3 Various special drive tasks 09.99 Without using a crawl velocity, a maximum positioning error, given by the following, would be obtained: smax = vmax t max = 36 m / min 4.8 ms = 0.29 cm ! Closed-loop frequency control is used as result of the enhanced control characteristics. Using closed-loop frequency control and the possibility of impressing the current at low frequencies, it is possible to obtain defined characteristics when ramping down from the crawl speed to 0. Closedloop speed control can also be used to prevent an excessive dependency on the load. 148 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks 3.3.3 Elevator drive with closed-loop positioning control (direct approach) A SIMOVERT VC drive converter with T300 technology board for positioning is used. The required accuracy when stopping at floors is 2.5 mm. Drive data Gearbox ratio i = 35 Drive wheel diameter D = 640 mm Max. velocity vmax = 1.25 m/s Max. acceleration amax = 0.9 m/s2 Checking the accuracy A double-pulse encoder with 1024 pulses per revolution is used. Thus, the smallest detectable position difference is given by: x di = D 640 = = 0.014 mm i 4 z 35 4 1024 This value is more than 400 % less than the required accuracy of 2.5 mm. Thus, the selected encoder is adequate as far as the resolution is concerned. The maximum motor speed is given by: n max = v max i 60 125 . 35 60 = 1305 RPM = D 0.64 For a speed control range of 1:1000 and a rated motor speed of nn=1500 RPM, the minimum velocity is given by: v min = nn D 1500 640 = = 144 . mm / s 1000 i 60 1000 35 60 If the closed-loop control should still respond to a position difference of xd = 2.5 mm = 0.625 mm 4 the following kv factor must be able to be set for the closed-loop position controller: kv = v min 144 . = = 2.3 s -1 0.625 xd This value can be realized for VC drive converters using T300. Siemens AG SIMOVERT MASTERDRIVES - Application Manual 149 3 Various special drive tasks 09.99 Velocity characteristic when moving between floors With stotal = h1 = 2.875 m this results in a positioning time for a floor of: t total = v max stotal 125 . 2.875 + = + = 3.96 s 0.9 125 . a max v max Accelerating- and decelerating time: tb = t v = v max 125 . . s = = 139 0.9 a max The constant velocity time is given by: t k = t total - 2 v max 2 125 . = 3.69 - = 0.91 s a max 0.9 Velocity characteristic when moving between floors v 1.25 m/s s total = 2.875 m 1.39 s 0.91 s 1.39 s t 3.69 s a 0.9 m/s 2 t - 0.9 m/s 2 150 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks 3.4 Drives with periodic load changes and surge loads 3.4.1 General information Periodic load changes and surge loads can occur, for example, for reciprocating compressor or press drives. For drives with periodic load changes, the following must be observed: * In order that the drive stability is guaranteed, a drive converter with vector control should be used. * If the excitation frequency of the load changes is higher than the limiting frequency of the closed-loop control (nominal values for closed-loop frequency control: Approx. 5 Hz, closedloop speed control: Approx. 15 Hz), then the closed-loop frequency/speed controller must be set as slow as possible, so that it does not attempt to compensate the speed fluctuations. * The motor torque to be generated is reduced by the drive moment of inertia, and for the extreme case, to the arithmetic average value of the load torque. This damping effect decreases as the speed decreases. * For drives with a low moment of inertia and high load torque changes (e. g. a single-cylinder reciprocating compressor), the motor cannot be simply dimensioned according to the RMS value of the load torque and the speed control range. In this case, the rated motor torque should be at least 80% of the maximum load torque. * When starting loaded (open-loop controlled range), under certain circumstances, the rated motor torque must be 130 to 150 % of the maximum load torque when starting. Vector control with tachometer is preferably used. In this case, it is sufficient when the rated motor torque approximately corresponds to the maximum load torque. * If the motor and load are elastically coupled (e. g. through a belt drive), then torsional oscillations can be generated. The excitation frequency of the load changes may not coincide with the resonant frequency. Drives with surge/shock loading: * If the drive moment of inertia is appropriately high, the surge torque can briefly be a multiple of the rated motor torque. * If there are only low moments of inertia, the motor can be braked down to standstill when load surges occur. Under certain circumstances, the drive can even briefly accelerate in the opposing direction of rotation (e. g. a driven vehicle collides with a stationary vehicle). In order to handle cases such as this, vector control with tachometer must be used in order to prevent the motor stalling. Siemens AG SIMOVERT MASTERDRIVES - Application Manual 151 3 Various special drive tasks 09.99 Supplementary moment of inertia By using a supplementary moment of inertia, it is possible to reduce the motor peak torques. When using a gearbox, it makes the most sense to locate this supplementary moment of inertia at the gearbox side with the higher speed. This results in smaller dimensions. Calculating the motor torque Under the assumption that the motor and load are rigidly coupled (no two-mass system with elastic coupling) and when neglecting the electrical motor time constants, the following differential equation is valid: Tmech dM motor + M motor = M load (t ) dt with Tmech = J total 2 n0 sn M n 60 mech. time constant [s] J total total moment of inertia [kgm2] Mn rated motor torque [Nm] sn = n0 - nn n0 rated motor slip n0 motor synchronous speed [RPM] nn motor rated speed [RPM] If the moment of inertia is known, for simple cases, the motor torque can be calculated. However, this calculation assumes that the drive is not in the closed-loop speed controlled mode. When using closed-loop speed control, depending on the setting, higher motor torques are obtained. In the extreme case, if the speed control could keep the speed precisely constant, the motor torque would precisely correspond to the load torque. 152 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks Case 1 The load torque is specified in the form of an arithmetic average value component with superimposed oscillation: M load (t ) = M load + M load sin( load t ) Thus, the motor torque is given by: M motor (t ) = M load + M load 1 + ( load Tmech ) 2 sin( load t - ) with = arctan( load Tmech ) The ratio of the amplitudes is then given by: M motor 1 = M load 1 + ( load Tmech ) 2 Thus, load torque oscillations are better damped the higher the mechanical time constant, and the higher the frequency of the load torque oscillation. Mload, M mot M load M mot M load load M mot M load 2 T= _ load t Example of load torque oscillation with load Tmech = 3 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 153 3 Various special drive tasks 09.99 Case 2 The load torque is a surge torque in the form of a step function. Thus, the motor torque is given by: M motor (t ) = M load max (1 - e - t Tmech 0 t t0 ) and M motor (t ) = M load max (1 - e - t0 Tmech )e - t -t0 Tmech t t0 M load, M mot M load max M load M M mot max t mot 0 Example for a load surge with Tmech = 3 t0 If Tmech is significantly longer than t0, the motor only has to provide a fraction of the maximum load torque. In this case, the maximum motor torque can be approximately calculated as follows: M motor max M load max 154 t0 Tmech Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks Case 3 The load torque is a periodic sequence of torque steps. Thus, the motor torque is given by: M motor (t ) = M load max + C1 e - t Tmech 0 t t0 and M motor (t ) = C2 e - t - t0 Tmech t0 t T With C1 = - M load max 1- e 1- e - - C2 = M load max + C1 e M M load, M tp Tmech T Tmech - t0 Tmech mot load max M M M M load M mot max mot load mot min t 0 tp t T Example of a periodic sequence of load steps with Tmech = 2.5 t 0 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 155 3 Various special drive tasks 09.99 If Tmech is significantly longer than to, the maximum and minimum motor torques can be approximately calculated as follows: M motor max M load + M load max t0 2 Tmech M motor min M load - M load max t0 2 Tmech with M load = M load max t0 T (arithmetic average value component of the load torque) Case 4 The load torque is specified as diagram. The following calculation procedures can be used for the motor torque: * The load torque characteristic can be approximated using simple functions, which can be solved using differential equations. * Fourier analysis of the load torque characteristic, and therefore effective as for case 1. In this case, in addition to the basic fundamental, only the harmonics have to be taken into account, which represent a significant proportion of the motor torque. * 156 Numerical integration, e. g., using the Runge Kutta technique Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks Example The load torque characteristic of a single-cylinder reciprocating compressor can be approximately represented as: Mload M load max Mload (approximation) M load 2 load t 2 corresponds to a crank revolution The following is valid for the approximation: M load (t ) = M load max sin( load t ) 0 load t M load (t ) = 0 load t 2 with load = 2 nmotor = 2 f load i 60 load torque oscillation frequency i gearbox ratio Thus, the motor torque is obtained from the following differential equation: M mot (t ) C e M mot (t ) C e - - load t load Tmech + M load max 1 + ( load Tmech ) 2 load t - load Tmech Siemens AG SIMOVERT MASTERDRIVES - Application Manual sin( load t - ) 0 load t load t 2 157 3 Various special drive tasks 09.99 with M C= (1 - e load max - load Tmech sin ) 1 + ( load Tmech ) 2 = arctan( load Tmech ) The motor torque is now to be represented for a control range of 1:4. load Tmech = 15 . for nmotor = nmotor n . For nmotor = nmotor n / 4 , a value of 0.375 is obtained. M load, M mot M load m max Mload (approximation) M mot M load 2 load t Motor torque at nmotor = nmotor n and load Tmech = 15 . Using the Mload characteristic,the arithmetic average value component of the load torque is approximately given by: M load M load max The calculation indicates that the maximum motor torque is only approximately 65 % of the maximum load torque. 158 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks M load, M mot Mload (approximation) M load max M mot M load Motor torque at nmotor = nmotor n 4 2 load t and load Tmech = 0.375 As a result of the low influence of the moment of inertia at low speeds, the maximum motor torque is almost as high as the maximum load torque. This approximation does not take into account that the load torque is also briefly negative. For low moments of inertia, or low speeds, the motor torque can also be negative (regenerative operation). Instead of directly solving a differential equation, the motor torque can also be calculated using Fourier analysis. The load torque is approximated by: M load (t ) 1 cos(2 load t ) 1 = M load max sin( load t ) + (1 - 2 ) 4 2 - 1 =1 2 2 2 1 1 M load max + sin( load t ) - sin( 2 load t + ) - sin( 4 load t + ) 3 2 15 2 2 (terminated after 4 terms of the progression) Siemens AG SIMOVERT MASTERDRIVES - Application Manual 159 3 Various special drive tasks 09.99 Thus, the motor torque is given by: 1 1 M motor (t ) M load max + sin( load t - 1 ) 2 1 + ( load Tmech ) 2 - - 2 3 1 + (2 load Tmech ) 2 2 15 1 + (4 load Tmech ) 2 sin(2 load t - 2 + ) 2 sin(4 load t - 4 + ) 2 with 1 = arctan( load Tmech ) 2 = arctan(2 load Tmech ) 4 = arctan(4 load Tmech ) The calculation method using 4 terms of the progression practically provides the same results as the previous direct calculation. 160 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks 3.4.2 Single-cylinder reciprocating compressor A single-cylinder reciprocating compressor is to be operated in a speed control range from 94 to 281 RPM. A gearbox with i = 5.33 is used. The load torque characteristic is known, and the load moment of inertia referred to the motor shaft. M *load / Nm 160 140 120 100 80 60 40 20 0 90 180 270 360 Crankshaft angle / degrees -20 Load torque characteristic of a single-cylinder reciprocating compressor referred to the motor shaft Drive data Max. load torque referred to the motor shaft M load max Min. load torque referred to the motor shaft M load min = -15 Nm Load moment of inertia referred to the motor shaft J load Max. motor speed nmotor max = 1500 RPM Min. motor speed nmotor min = 500 RPM = 0.0213 kgm2 As there is no significant moment of inertia, a motor is required with a rated torque of at least 80% of the maximum load torque referred to the motor shaft. M motor n 0.8 M load max = 0.8 153 = 122.4 Nm Siemens AG SIMOVERT MASTERDRIVES - Application Manual 161 3 Various special drive tasks 09.99 A 22 kW, 4-pole motor is selected (1LA5 186-4AA..) Mn =144 Nm, nn =1460 RPM, Imotor n = 41 A, Jmotor = 0.15 kgm2, motor = 0.912 Estimating the influence of the moment of inertia on the maximum motor torque Total moment of inertia referred to the motor shaft: Jtotal = J motor + J load = 015 . + 0.0213 = 01713 . kgm2 Rated motor slip: sn = n0 - nn 1500 - 1460 = = 0.0267 n0 1500 Mechanical time constant: Tmech = Jtotal 2 n0 sn 01713 2 1500 0.0267 . = = 0.005 s 144 60 M n 60 Excitation frequency at nmotor max = 1500 RPM: load = 2 nmotor max i 60 = 2 1500 = 29.47 s -1 5.33 60 or f load = load 29.47 = = 4.69 Hz 2 2 Thus, the following is obtained for load Tmech at nmotor max = 1500 RPM: load Tmech = 29.47 0.005 = 0147 . The basic fundamental amplitude ratio is then given by: 1 1 M motor = = = 0.99 2 M load 1 + ( load Tmech ) 1 + 0147 . 2 The damping effect of the moment of inertia can in this case, even at maximum speed, be neglected for the basic fundamental. This means, that the maximum motor torque, mainly determined by the basic fundamental, is practically the same as the maximum load torque referred to the motor shaft. 162 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks Estimating the braking energy M*load ~ ~ 27.5 180 Crankshaft angle/degrees 210 -15 Nm Section of the load torque characteristic The motor tracks the positive section of the load torque characteristic practically 1:1, with the exception of a phase shift. In the negative section, the peak of the load torque is better damped as a result of the higher frequency. In order to estimate the maximum negative motor torque, a sinusoidal oscillation with 180/27.5 = 654 % frequency is assumed. The approximate motor torque at the maximum motor speed is given by: M motor M load 1 + (6.54 load Tmech ) 2 = 15 1 + (6.54 0147 . )2 = 10.8 Nm The maximum braking power is given by: Pbr max M motor nmotor max motor 9550 = 10.8 1500 0.912 = 155 . kW 9550 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 163 3 Various special drive tasks 09.99 Braking diagram Pbr ~ ~ Pbr max 180 210 Crankshaft angle/degrees The braking energy of a cycle corresponds to the area in the braking diagram. The braking characteristic is given by: Pbr Pbr max sin(6.54 load t ) Thus, the braking energy at nmotor max = 1500 RPM is given by: Wbr 2 Pbr max 6.54 load = 2 155 . = 0.0161 kWs = 161 . Ws 6.54 29.47 It should now be checked, as to whether the drive converter DC link can accept this energy. The maximum amount of energy which can be accepted by the DC link for a 22 kW drive converter on the 400 V line supply is as follows: Wmax = 1 1 C (Vd2max - Vd2n ) = 2700 10 - 6 (820 2 - 540 2 ) = 514.1 Ws 2 2 Thus, the braking energy, which occurs during the brief regenerative phase, can be temporarily accepted by the DC link. It is then reduced to zero in the subsequent motor operation. Thus, a pulsed resistor is not necessary. Drive converter selection 6SE7024-7ED61 PV n = 22 kW, IV n = 47 A Closed-loop frequency control 164 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks Thermally checking the motor As the motor torque practically corresponds to the load torque referred to the motor shaft, when calculating the RMS motor torque, the RMS load torque can be used as basis. The load torque is approximated using a sinusoidal half wave (as described previously under "General information"). M*load max 0 90 180 270 360 Crankshaft angle/degrees Approximation of the load torque characteristic The RMS motor torque is approximately given by: M motor RMS M load max 2 = 153 = 76.5 Nm 2 In a speed control range 1:3 for the 22 kW motor, the permissible torque for S1 duty is approximately 111 Nm. Thus, there is sufficient reserve. Starting under load If a single-cylinder reciprocating compressor is to start under load, for closed-loop frequency control, a 30 kW motor and a 30 kW drive converter are required (6SE7026-0ED61). Siemens AG SIMOVERT MASTERDRIVES - Application Manual 165 3 Various special drive tasks 09.99 Motor torque calculated using the Runge Kutta technique R e c ip ro c a tin g c o m p re s s o r (n = 1 5 0 0 R P M , i= 5 .3 3 , T m e c h = 0 .0 0 5 s ) 1 60 1 40 Torque in Nm 1 20 1 00 Load torqu e 80 M oto r torque 60 A v erage v alue 40 20 0 -2 0 0 30 60 90 120 150 18 0 210 240 27 0 300 330 36 0 C ra nks ha ft a ng le in d e g re e s 166 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks 3.4.3 Three-cylinder pump A three-cylinder pump is to be operated in a control range 102 to 238 RPM. A gearbox with i = 6.11 and = 0.97 is used. The load torque characteristic is specified and the torque at the pump shaft is 796 Nm. M load / M load 1,4 1,235 1,2 1 0,8 0,686 0,6 0,4 0,2 0 0 30 60 90 120 150 180 210 240 270 300 330 360 Crankshaft angle/degrees Load torque characteristic of the three-cylinder pump With the average torque at the pump shaft and a gearbox efficiency of 0.97, the average torque at the motor shaft is obtained as follows: M motor = M load 796 = = 134.3 Nm i 611 . 0.97 The load moment of inertia is not known, but can probably be neglected with respect to the motor moment of inertia. Thus, it is assumed that the motor torque has the same ripple as the load torque. The maximum and minimum motor torques are given by: M motor max = 1235 . M motor = 1235 . 134.3 = 165.9 Nm M motor min = 0.686 M motor = 0.686 134.3 = 92.1 Nm Siemens AG SIMOVERT MASTERDRIVES - Application Manual 167 3 Various special drive tasks 09.99 The maximum and minimum motor speeds are given by: nmotor max = nload max i = 238 611 . = 1454.2 RPM nmotor min = nload min i = 102 611 . = 623.2 RPM As result of the relatively low load torque oscillations, the RMS motor torque and the RMS load torque can be used to dimension and select the motor. The load torque characteristic is approximated using individual sinusoidal half waves to estimate the RMS value. M load / M load a1 1 a2 a3 a4 1 2 3 4 120 in degrees The load characteristic is approximated using sinusoidal half waves The approximate RMS value is obtained as follows: i =4 M load RMS = M load 4 ai ai2 + ) 2 120 i (1 + i =1 With a1 = 0.235, a2 = -0.061, a3 = 0.093, a4 = -0.3 1 = 44, 2 = 11, 3 = 24, 4 = 41 1.0111 is obtained as the square root. Thus, the RMS motor torque is approximated as follows: 168 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks M motor RMS M motor 10111 . = 134.3 1.0111 = 1358 . Nm If the arithmetic average value is high with respect to the oscillating torque amplitudes, then the RMS value practically corresponds to the average value. A 4-pole, 30 kW motor is selected due to the speed control range 1:2.33 (1LA5 207-4AA..) Mn = 196 Nm, nn = 1465 RPM, Imotor n = 55 A, Jmotor = 0.24 kgm2 Estimating the influence of the moment of inertia on the maximum motor torque Total moment of inertia referred to the motor shaft: Jtotal J motor = 0.24 kgm2 Rated motor slip: sn = n0 - nn 1500 - 1465 = = 0.0233 n0 1500 Mechanical time constant: Tmech = Jtotal 2 n0 sn 0.24 2 1500 0.0233 = = 0.0045 s M n 60 196 60 Excitation frequency at nmotor max = 1454.2 RPM: load = 3 2 nmotor max i 60 = 3 2 1454.2 = 74.8 s -1 611 . 60 (3 periods for a crankshaft revolution) and f load = load 74.8 = = 11.9 Hz 2 2 Thus, load Tmech at nmotor max = 1454.2 RPM is given by: load Tmech = 74.8 0.0045 = 0.336 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 169 3 Various special drive tasks 09.99 The basic fundamental amplitude ratio is now given by: 1 1 M motor = = = 0.95 2 M load 1 + ( load Tmech ) 1 + 0.3362 Thus, the damping effect of the moment of inertia, even at maximum speed can be neglected for the basic fundamental. Selected drive converter 6SE7027-2ED61 PV n = 37 kW, IV n = 72 A Closed-loop frequency control type Motor torque numerically calculated using the Runge Kutta technique T hre e -c ylind e r p um p (n=1 4 5 4 R P M , i=6 .1 1 , T m e c h= 0.0 04 5 s) 1,4 Torque referred to the average value 1,2 1 Load torqu e 0,8 M oto r torque 0,6 A v erage v alue 0,4 0,2 0 0 30 60 90 120 150 180 210 2 40 270 300 33 0 360 C ran ks h aft an g le in d eg rees 170 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3.4.4 3 Various special drive tasks Transport system for freight wagons This drive is for a trolley that pushes freight wagons to form train units on a 500 m rail section. The vehicle is driven using a fixed 37 kW AC motor with gearbox. Trolley Wagon v Cable ~ ~ ~ ~ Driving pulley Gearbox Approx. 500 m rail section IM Schematic of the trolley system! For a 50 Hz motor frequency, the trolley runs at a constant velocity of 1.25 m/s towards the wagons. The cable tension is established after a delay due to the spring effect of the steel cable. If the wagons are braked, or if a group of several wagons doesn't accelerate quickly enough due to the high mass, then a slip coupling between the motor and gearbox is activated when a maximum motor torque is reached. The pulley wheel speed decreases. If the wagon group doesn't move or the pulley wheel speed does not increase after a specific time, the trolley is retracted. In order to eliminate the slip coupling, the motor is now to be fed from a drive converter. Further, in order to save time, the trolley can be withdrawn at twice the velocity. The slip coupling is now replaced by a closed-loop speed/torque control. If the motor torque, transferred via the cable, reaches the set torque limit, the closed-loop speed control can no longer maintain the reference speed, and the drive system automatically goes over into the closed-loop torque controlled mode. The motor maintains the torque, the speed decreases, and is now defined by the load. However, without the damping effect of the slip coupling, problems can occur as the motor-cable-wagon combination represents a system which can oscillate. This will now be shown using the following simplified diagram. The friction and gearbox efficiency have, in this case, been neglected. Siemens AG SIMOVERT MASTERDRIVES - Application Manual 171 3 Various special drive tasks 09.99 v1 c v2 m i r IM x1 x2 Equivalent diagram for system behavior after the trolley has hit" the wagon x1 cable motion over the drive wheel x2 wagon motion v1 velocity of the cable over the drive wheel v2 velocity of the wagon(s) r drive wheel radius c spring constant of the cable (dependent on the cable length) m mass of the wagon(s) i gearbox ratio Jmotor motor moment of inertia n0 motor no-load speed at rated frequency The following is valid for the drive wheel: load F load r i M motor i M motor - M load = i 2 J motor 172 d load dt Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks with M load = Fload r = c ( x1 - x2 ) r load = v1 1 dx1 = r r dt The following is valid for the wagon(s): v2 m F tension Ftension = m dv 2 d 2 x2 = m dt dt 2 with Ftension = c ( x1 - x 2 ) Behavior when a wagon is braked To start of with, the behavior is investigated when the trolley hits a braked wagon. In this case, v2=0 and x2=0. It is now assumed, that the speed remains constant due to the closed-loop speed control, until Mmotor=Mmax. Thus, the following is true for this range: load = const = v10 2 n0 1 dx1 = = r r dt i 60 Solution: x1 = v10 t M motor = c r v10 t i Siemens AG SIMOVERT MASTERDRIVES - Application Manual 173 3 Various special drive tasks 09.99 Mmotor=Mmax is reached at t1 = i M max c r v10 After the torque limit has been reached, Mmotor=Mmax remains constant. The following is valid for this range: d 2 x1 c r2 i M r + x1 = 2 max 2 2 dt i Jmotor i J motor Solution: x1 = v10 i M max sin( m1 ( t - t1 )) + m1 cr dx1 = v1 = v10 cos( m1 (t - t1 )) dt with m1 = c r2 i 2 J motor mechanical oscillation frequency For operation at the torque limit, undamped oscillation occurs. However, in practice there is always some damping due to the cable friction and gearbox losses, which in this particular analysis, have been neglected. Additional damping can be achieved by compensating the speed to the torque limit. This compensation has, in this case, the following form: M motor = M max - k D M max n v = M max - k D M max 1 n0 v10 For example, for kD=0.1, a 10 % change in the torque limit is achieved at n=n0. With this compensation, the differential equation now has a damping component: d 2 x1 i k D M max r dx1 c r2 i M r + + 2 x1 = 2 max 2 2 dt v10 i J motor dt i J motor i J motor 174 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks If, for example, the start of compensation is set to Mmotor=0.2 Mmax, then, for non-periodic damping, the solution is as follows: x1 = C1 e p1(t -t2 ) + C2 e p2 (t -t2 ) + i M max cr dx1 = v1 = C1 p1 e p1(t -t2 ) + C2 p2 e p2 (t -t2 ) dt with p1,2 = - i k D M max r v10 i 2 J motor a1 = i M max ) - v10 cr p2 - p1 p2 ( v10 t2 - C1 = v10 - p1 ( v10 t2 - C2 = t2 = a1 a ( 1 )2 - 2m1 2 2 p2 - p1 i M max ) cr 0, 2 M max i c r v10 With the following values r = 0.36 m c = 20000 N/m i = 22.5 Jmotor = 1.2 kgm2 Mn = 483 Nm n0 = 750 RPM when the trolley hits the braked wagons and for Mmax=Mn, the following characteristics are obtained for the cable motion along the drive wheel x1, velocity of the cable along driving wheel v1, the motor torque Mmotor and tension Ftension. In order to keep the equation transparent, the quantities are referred to the relevant nominal values: x10 = i Mn cr and Fn = c x10 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 175 3 Various special drive tasks 09.99 Cable velocity across the drive wheel (v1), Cable route along the drive wheel (x1) kD=0 1,5 1 0,5 x 1/x 1 0 v 1/v 10 0 0 1 2 3 4 5 6 7 -0,5 -1 Time in s Motor torque (M Mot), tension (F tension) kD=0 1,6 1,4 1,2 1 M M o t/M n 0,8 F tension/F n 0,6 0,4 0,2 0 0 1 2 3 4 5 6 7 Time in s Behavior when the trolley hits the wagon(s) which is(are) braked. Without compensation, the system continuously oscillates (kD=0). 176 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks Cable velocity across the drive wheel (v1), Cable route along the drive wheel (x1) kD=1,4 1,5 1 x 1/x 1 0 v 1/v 10 0,5 0 0 1 2 3 4 5 6 7 Time in s Motor torque (M Mot), tension (F tension) 1,6 1,4 1,2 1 0,8 0,6 0,4 0,2 0 -0,2 0 -0,4 kD=1,4 F tension/F n M M o t/M n 1 2 3 4 5 6 7 Time in s Behavior when the vehicle impacts wagon(s) which is(are) braked. With the appropriate compensation (e. g. kD=1.4) a non-periodic damped oscillation is obtained. Compensation starts at Mmotor=0.2 Mmax. Siemens AG SIMOVERT MASTERDRIVES - Application Manual 177 3 Various special drive tasks 09.99 Behavior when accelerating the wagon(s) To start of with, the acceleration of an unbraked group of wagons will be investigated. As v2 and x2 now also change, the analytical calculation is significantly more complex. The equations are now changed-over as follows: dx1 = v1 dt dx2 = v2 dt dv2 c = ( x1 - x2 ) dt m as well as dv1 =0 dt without compensation and for v1 = v10 dv1 r i v1 cr2 = ( M max - k D M max ) - 2 ( x1 - x2 ) dt i 2 J motor v10 i J motor with compensation This equation system can be solved using the Runge Kutta technique. Using the same numerical values as before for braked wagons, Mmax=Mn, and mass m of 400 t or 10 t, the following characteristics are obtained for the velocity of the cable over drive wheel v1, the velocity of the wagon(s) v2, motor torque Mmotor and tension Ftension. Motor limit compensation is again withdrawn after approx. 1.5 s using a PT1 element with a time constant of approx. 5 s; otherwise, the motor torque during acceleration would be permanently reduced by factor kD. The overshoot of velocity v2 of the wagon(s) is certainly lower as friction has been neglected. This effect can be significantly reduced, for example, by decreasing the setpoint to approx. 80% after the vehicle has hit the wagon. If the motor speed again reaches this 80%, the setpoint is slowly increased to 100% along a slow ramp (ramp-function generator). Thus, the cable tension (and also overshoot) is reduced before the final velocity is reached. 178 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks Cable velocity across the drive wheel (v1), velocity of the wagons (v2) m=400 t, kD=1,4 1,4 1,2 1 0,8 v 1/v 10 0,6 v 2/v 10 0,4 0,2 0 0 5 10 15 20 25 30 Time in s Motor torque (M Mot), tension (F tension) m=400 t, kD=1,4 1 0,8 0,6 0,4 M M o t/M n 0,2 F tension/F n 0 -0,2 0 5 10 15 20 25 30 -0,4 Time in s Behavior when accelerating the wagons with m=400 t and a compensation of kD=1.4. Siemens AG SIMOVERT MASTERDRIVES - Application Manual 179 3 Various special drive tasks 09.99 Cable velocity across the drive wheel (v1), velocity of the wagons (v2) m=10 t, kD=1,4 1,2 1 0,8 v 1/v 10 0,6 v 2/v 10 0,4 0,2 0 0 0,5 1 1,5 2 2,5 3 Time in s Motor torque (M Mot), tension (F tension) m=10 t, kD=1,4 0,4 0,3 0,2 0,1 M M o t/M n 0 -0,1 0 0,5 1 1,5 2 2,5 3 F tension/F n -0,2 -0,3 -0,4 Time in s Behavior when accelerating a wagon with m=10 t and a compensation of kD=1.4. 180 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks Implementing the torque compensation function n kD M n _ n0 n Speed controller S1 Mn n* m* T=5s S2 kD Mn n n 0 Switch S2 is closed before impact. Switch S1 closes after impact at Mmotor=0.2 Mmax. S2 opens after 1.5 s and ensures that compensation is slowly withdrawn. This can be realized, for example, using the T100 technology board. Selecting the motor and drive converter The motor already exists. Type: ABB with In=81 A; nn=735 RPM; Jmotor=1.2 kgm2; =0.93; Mn=483 Nm; Mstall/Mn=2.2 Drive converter selected: 6SE7031-0EE60 PV n=45 kW; IV n=92 A; IV max=125.6 A Closed-loop speed control T100 technology board Dimensioning the brake resistor Braking energy is generated when: a) the trolley brakes to 0 after it moves backwards at 2.5 m/s (twice the velocity) Siemens AG SIMOVERT MASTERDRIVES - Application Manual 181 3 Various special drive tasks 09.99 b) the trolley brakes from 1.25 m/s to 0 while moving forwards c) during impact Case a) When braking from twice the velocity in the speed range between 1500 RPM and 750 RPM, the motor is operated in the field-weakening range. Permissible motor torque at nmax and 130 % safety margin from the stall torque: M perm. = M stall n M stall n 2.2 483 750 2 = ( 0 )2 = ( ) = 204.4 Nm 1.3 13 . 13 . 1500 n max For Mmotor=Mperm=constant while braking, then the maximum braking power is given by: Pbr max = M perm. n max 9550 motor = 204.4 1500 0.93 = 29.9 kW 9550 The load moment of inertia is required to determine the braking time. The linearly-moved masses (trolley and cable) are assumed to be 1000 kg. Then, the following is true: Jload = m r 2 = 1000 0.362 = 129.6 kgm2 or, referred to the motor * = Jload Jload 129.6 = = 0.256 kgm2 2 2 22.5 i Thus, when neglecting the friction and gearbox losses, the braking time is given by: t br = * ( J motor + J load ) 2 n max (12 . + 0.256) 2 1500 = = 112 . s 60 M perm. 60 204.4 Case b) Braking from 750 RPM to 0 in the constant-flux range is to be realized in, for example, 0.5 s. Thus, the motor torque is given by: M motor = * ( Jmotor + Jload ) 2 n0 (12 . + 0.256) 2 750 = = 228.7 Nm 60 tbr 60 0.5 Thus, the maximum braking power is given by: Pbr max = 182 228.7 750 M motor n0 motor = 0.93 = 16.7 kW 9550 9550 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks Case c) As can be seen from the diagrams, the most unfavorable case is when the wagon(s) is(are) braked without torque compensation. The theoretical maximum braking power for Mmotor=Mn and n=-n0 is obtained as follows: Pbr max = M n n0 483 750 motor = 0.93 = 35.3 kW 9550 9550 However, this value does not actually occur in practice as the cable friction and gearbox losses were neglected. The behavior with a damping compensation of kD=1.4 is more realistic. In this case, the maximum braking power after impact with Mmotor=-0.4 Mn and n=n0 is given by: Pbr max = 0.4 M n n0 0.4 483 750 motor = 0.93 = 14.1 kW 9550 9550 Thus, a maximum braking power for case a) of 29.9 kW is obtained. A braking unit with P20=20 kW (6SE7023-2EA87-2DA0) and external brake resistor (6SE7023-2ES87-2DC0) would be sufficient. This allows a peak braking power of 1.5 P20=30 kW to be achieved for 3 s. Siemens AG SIMOVERT MASTERDRIVES - Application Manual 183 3 Various special drive tasks 3.4.5 09.99 Drive for an eccentric press A press drive with a 160 kW DC motor and nn = 2400 RPM is to be replaced by a three-phase drive. The DC motor speed control range is 436 RPM (setting-up speed) up to 2400 RPM (maximum speed). Flywheel DM z=19 DS z=64 z=31 z=170 r l Simplified function schematic Drive data Max. pressing force 25 mm before lower end stop F load max = 10000 kN Max. deforming work W max = 250 kWs Diameter of the motor pulley wheel DM = 300 mm Diameter of the drive wheel DS = 1670 mm Moment of inertia of the flywheel JS = 2845 kgm2 Crank radius r = 470 mm 184 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks Effective crank angle during impact St = 18.8 degrees Number of strokes at 436 RPM (crawl speed nmin) zmin = 4/min Number of strokes at 873 RPM (base speed nG) zG = 8/min Number of strokes at 2400 RPM (max. speed nmax) zmax = 22/min Required accelerating time from 0 to nmax th < 60 s Required braking (deceleration) time from nmax to 0 tbr < 60 s Using this drive data, it has to be checked whether a 4-pole 160 kW induction motor is suitable for the press drive. Motor data 1LA6 316-4AA; nn=1485 RPM; Mn=1030 Nm; Jmotor=3.2 kgm2; motor=0.958; Mstall=2.7 Mn; Imotor n=275 A The average motor power at maximum speed and an assumed press efficiency of 0.8: Pmotor average = Wmax z max 250 kWs 22 RPM 250 22 = = = 114.6 kW press press 0.8 60 As far as the average power is concerned, this motor is adequate. However, the behavior during impact as well as the RMS motor loading should now be investigated. Calculating the ratios The ratio between the motor and gearbox is obtained as follows: iS = DS 1670 = = 5.57 DM 300 The total ratio between the motor and the crank wheel is obtained from the number of teeth as follows: itotal = 1670 64 170 = 102.83 300 19 31 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 185 3 Various special drive tasks 09.99 Calculating the load torque In the following calculation, the press efficiency, additional frictional forces and the moment of inertia of the gear are neglected. Further, it is assumed that the load force is constant during impact. load r FT F T: Tangential force h: Ram displacement during pressing l h F load max M load = FT r = Fload max r sin( + ) cos h = r (1 - cos St ) For l/r >> 1, or for a small angle , the equation for Mload is simplified as follows: M load = FT r Fload max r sin For =St=18.8 degrees, the maximum load torque referred to the motor is obtained as follows: * M load max = 186 Fload max r sin St itotal = 10000 kN 0.47 sin( 18.8 102.83 ) 180 = 14.73 kNm Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks The motor cannot produce this torque. It must be provided by the flywheel. Calculating the mechanical time constant: Tmech = J * 2 nn s n M n 60 With J* = sn = JS 2845 + Jmotor = + 3.2 = 94.9 kgm2 2 iS 5.572 nsyn - nn 1500 - 1485 = = 0.01 nsyn 1500 (total moment of inertia referred to the motor) (motor slip) the following is obtained: Tmech = 94.9 2 1485 0.01 = 0.143 s 1030 60 The worst case condition, with impact at the lowest operating speed, is now investigated i.e. at the base speed. The time for the impact at nG =873 RPM, but neglecting the speed dip, is obtained as follows: t St itotal 60 = = load 2 nG 18.8 102.83 60 180 = 0.369 s 2 873 As the following is not valid for the mechanical time constant: Tmech >> tSt, the motor quickly reaches the torque limit. The motor speed is then mainly defined by the load. In order to simplify the calculation, it is assumed that the torque is limited at the start of impact. Then, the following is true: itotal M motor max - Fload max r sin = J load load = - d load dt d dt Siemens AG SIMOVERT MASTERDRIVES - Application Manual 187 3 Various special drive tasks 09.99 With 2 J load = J * itotal M motor max = torque limit This differential equation system can be solved, for example, using the Runge Kutta technique. For a small angle , with sin , a differential equation can be analytically solved as follows: itotal M motor max d 2 Fload max r - = - 2 dt J load J load Solution: = C1 e p t + C2 e p t + 1 load = - 2 itotal M motor max Fload max r d = - C1 p1 e p1t - C2 p2 e p2t dt nmotor = itotal load 60 2 with p1,2 = Fload max r J load C1 = itotal M motor max 2 nmotor 1 1 ( St - - ) 2 Fload max r p1 60 itotal C2 = itotal M motor max 2 nmotor 1 1 ( St - + ) 2 Fload max r p1 60 itotal With the specified data, and the specification M motor max = 15 . M n , the characteristic of load and nmotor can be calculated during the impact. For nmotor 1=nG (motor speed before impact), by iterating for =0, the following values are obtained: nmotor 2 = 613 RPM (motor speed after the impact) t St = 0.47 s (impact time) 188 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks Towards the end of the impact, if the motor torque exceeds the load torque, the flywheel is again accelerated with M motor = M motor max . The following is valid: itotal M motor max = J load d load dt Thus, the motor speed is given by: nmotor = nmotor 2 + 2 itotal M motor max 60 J load 2 (t - t St ) or for the accelerating time from nmotor 2 to nmotor 1: J * 2 tb = = M motor max 60 (nmotor 1 - nmotor 2 ) 94.9 2 (873 - 613) = 167 . s 15 . 1030 60 The time from the start of impact until the original speed has been reached again, is given by: t St + tb = 0.47 + 167 . = 2.14 s The maximum available time at nG, with 8 strokes/min is obtained as follows: t max = 1 min 60 s = = = 7.5 s zG 8 8 The load torque and the motor speed at the lowest operating speed nmotor 1=nG=873 RPM is illustrated in the following diagrams. Siemens AG SIMOVERT MASTERDRIVES - Application Manual 189 3 Various special drive tasks 09.99 M load / kNm 1600 1200 800 400 0 0 0.5 1 1.5 2 2.5 t/s nmotor / RPM 1000 nmotor 1 800 n motor 2 600 t St tb 400 200 0 0 0.5 1 1.5 2 2.5 t/s Load torque and motor speed at nmotor 1=nG=873 RPM 190 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks Simplified calculation Using the conservation of energy equation, the magnitude of the speed dip as well as the impact time can be approximately calculated. The energy supplied from the flywheel is: W = 1 * 2 2 2 2 J ( ) (nmotor 1 - nmotor 2 ) = Wmax - Wmotor 2 60 W motor is the energy supplied from the motor during impact: St Wmotor = M motor itotal d = M motor max itotal St 0 Thus, the motor speed after the impact is given by: 2 nmotor 2 = nmotor 1- Wmax - M motor max itotal St 1 * 2 2 J ( ) 2 60 The approximate impact time is given by: t St itotal St motor with the average motor speed: motor = 2 nmotor 1 + nmotor 2 60 2 Using these formulas, the speed dip and the impact time for nmotor 1=873 RPM are again checked. nmotor 2 = 8732 - 250 103 - 15 . 1030 102.83 18.8 1 2 94.9 ( ) 2 2 60 180 = 618 RPM 180 = 0.43 s t St 2 873 + 618 60 2 102.83 18.8 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 191 3 Various special drive tasks 09.99 Behavior in the field-weakening range The drive is to be operated with constant power in the field-weakening range. Further, there should be a 130 % safety margin to the stall limit. Thus, the permissible motor torque is represented by the following characteristics. M motor perm M stall 1,3 P = const nn n max n motor If a torque of 1.5 Mn is assumed in the constant-flux range, then the permissible torque at constant power for nmax is obtained as follows: M perm. ( nmax ) = nn 1485 15 . Mn = 15 . 1030 = 966 Nm n max 2400 On the other hand, with Mstall=2.7 Mn, for nmax, the permissible torque for a 130 % safety margin is obtained as follows: M perm. ( nmax ) = M stall n 2.7 M n 1485 2 2.7 1030 = ( n )2 =( ) = 819 Nm 1.3 n max 13 . 2400 13 . Thus, the limit for Mperm.(nmax) is defined by the stall torque. With Mperm.(nmax), the speed dip and the impact time can be calculated for nmotor=nmax=2400 RPM. nmotor 2 = 24002 - 192 250 103 - 819 102.83 18.8 1 2 94.9 ( ) 2 2 60 180 = 2309 RPM Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks 180 = 0137 . t St s 2 2400 + 2309 60 2 102.83 18.8 The following is valid for the accelerating time from nmotor 2 to nmotor 1: tb = J * 2 M motor perm. ( nmax ) 60 (nmotor 1 - nmotor 2 ) = 94.9 2 . s (2400 - 2309) = 11 819 60 The condition tSt + tb < t max = 124 . s < 1 z max 60 s = 2.73 s 22 is also fulfilled. Calculating the accelerating- and decelerating times If the drive is accelerated from 0 to n max, with the torque permissible at nmotor = nmax in the field weakening range, or is braked from n max to 0, the accelerating- and decelerating times are given by: t h = t br = J * 2 nmax 94.9 2 2400 = = 29.12 s 60 M motor perm.( nmax ) 60 819 This time is less than the specified 60 s. Siemens AG SIMOVERT MASTERDRIVES - Application Manual 193 3 Various special drive tasks 09.99 Thermally checking the motor Torque characteristic M motor M motor max t t St+tb T The RMS torque is obtained from the torque characteristic: M RMS = M motor max t St + tb T With T= 1 z In this case, factor kf is 1, as it involves continuous operation. The most unfavorable case is obtained at the lowest operating speed nG, as in this case, the speed dip, and thus also the ratio tSt+tb/T is its highest (at nmax, the RMS torque is always less than Mn due to Mmotor max < Mn). For nmotor=nG, the following value is obtained for MRMS: M RMS = 15 . Mn t St + tb 2.14 = 15 . 1030 = 825 Nm 1 7.5 zG Thus, the calculated RMS torque is less than the permissible motor torque in the speed control range 1:2. Thus, operation is thermally permissible. 194 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks Selecting the drive converter At 150 % motor torque, the maximum motor current is given by: 2 2 2 I motor max 15 . 2 ( I motor n - I n ) + I n With I n = 0.27 I motor n , the following is obtained: I motor max 15 . 2 (2752 - 0.27 2 2752 ) + 0.272 2752 = 404 A Thus, the RMS motor current can be calculated for the most unfavorable case at nmotor=nG. I motor I motor max Imag n t t St + t b T From the diagram above, the following is obtained for Imotor RMS: I motor RMS = = 2 2 I motor max ( t St + tb ) + I n ( T - ( t St + tb )) T 404 2 2.14 + 0.27 2 2752 (7.5 - 2.14) = 225 A 7.5 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 195 3 Various special drive tasks 09.99 Selected drive converter: 6SE7033-2EG60 PV n=160 kW; IV n=315 A; IV max=430 A Closed-loop frequency control mode The rated drive converter current is greater than the RMS motor current. The maximum motor current therefore does not exceed the maximum drive converter current. Dimensioning the brake resistor If the drive is also braked with the permissible torque at nmax in the field-weakening range, the maximum braking power for the brake resistor is obtained as follows: Pbr W max = M motor perm. ( nmax ) n max 9550 motor = 819 2400 0.958 = 197.2 kW 9550 The braking energy when braking from nmax to 0 at tbr=th is then as follows: Wbr = 1 1 Pbr W max tbr = 197.2 29.12 = 2871 kWs 2 2 The following must then be valid for the brake resistor: Wbr 2871 = = 319 . kW Pbr cont . 90 T With Pbr cont . = P20 4.5 (with an external brake resistor) the following is obtained 319 . 4.5 = 143.6 kW P20 Thus, a braking unit is selected with P20 = 170 kW (6SE7032-7EB87-2DA0) with an external brake resistor (6SE7032-7ES87-2DC0). 196 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks 3.5 Load distribution for mechanically-coupled drives 3.5.1 General information For mechanically-coupled drives, for example traction drives with several driven axes, or roller table drives which are coupling through the material, it is important to uniformly distribute the load over the individual drive motors. Depending on the particular application case, this can be implemented by using a group (sectional) drive with one drive converter, or using individual drives, each with its own dedicated drive converter. Ring gear IM IM Pinion Pinion Example of a mechanically-coupled drive Multi-motor drive configuration The simplest way of achieving a uniform load distribution for mechanically-coupled drives with motors having the same output is by using a multi-motor drive configuration. In this case, all of the motors are connected to a drive converter, and are therefore supplied with the same stator frequency. As a result of the mechanical coupling, the motors have the same speed with symmetrical relationships (identical gearbox ratio, identical roll diameter etc.). Deviations in the load distribution can therefore only come if the individual motors have different slip characteristics. In accordance with DIN VDE 0530, deviations of up to 20% are permissible. A specially-selected motor can also help. Problems can also occur for traction units with slightly different drive wheel diameters. The thus obtained speed differences increasingly influence the load distribution the lower the rated motor slip is. Siemens AG SIMOVERT MASTERDRIVES - Application Manual 197 3 Various special drive tasks 09.99 n n2 n1 M2 M total M1 M Example for the load distribution of 2 mechanically-coupled motors with different slip characteristics (the motor with the flatter characteristic accepts the higher load component) n n2 n M2 M total M1 n 1 M Example for the load distribution of 2 mechanically-coupled motors with different speeds, e. g. as a result of different wheel diameters (the motor with the lowest speed accepts the higher load component) Group drives cannot be used with different motor outputs to provide a uniform load distribution if technological requirements oppose this (e. g. redundancy), or if the speed differences are too high. 198 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks Single-motor drive configuration For single-motor drives with motors having the same output, when the same setpoint is entered at all of the drive converters, characteristics can be achieved which are similar to those for a group drive. The setpoints should be digitally entered so that the stator frequencies, are, as far as possible, identical (e. g. via a peer-to-peer connection of the individual drive converters). Further, the same ramp-up and ramp-down times must be selected for all of the drive converters. When using the V/f characteristic or closed-loop frequency control, slip compensation must be disabled. Otherwise, load distribution cannot be realized through the motor slip. It is also not possible to use closed-loop speed control for the same reason (i.e. without "droop"). Further, it is possible to influence the load distribution by activating the so-called "droop" characteristic. Droop characteristic SIMOVERT VC drive converters must be used when implementing the droop function. The droop function can be used for closed-loop frequency- or for closed-loop speed control. For closed-loop frequency control, it is only effective in the controlled range (from approx. 5 Hz). Kp n*/f * m* Speed/frequency controller n act /f act Block diagram of "droop" for SIMOVERT VC The droop function causes the motor slip to increase when a load is applied to the drive due to the negative feedback of the torque setpoint to the speed / frequency setpoint. Thus, under steadystate conditions, uniform load distribution can be achieved by setting the same slip characteristic. "Load equalization control" and "master/slave drive" techniques provide a flexible way of ensuring unified load distribution, even at various speeds. Siemens AG SIMOVERT MASTERDRIVES - Application Manual 199 3 Various special drive tasks 09.99 Load equalization control SIMOVERT VC drive converters must be used when implementing load equalization control. Closed-loop frequency control must be set. All of the drive converters receive the same frequency setpoint, and the same ramp-up and ramp-down times must be set. A uniform load distribution is achieved by comparing the torque setpoint of the first drive with the torque setpoints of all other drives, and then a frequency correction setpoint for these drives is generated by the PI controller, provided as standard in SIMOVERT VC drive converters. This correction setpoint must be limited to approx. 300 to 400 % of the motor rated slip frequency. The load equalization control may only be enabled, if the closed-loop frequency control is fully operational (from approx. 5 Hz). Below this range, the open-loop controlled mode is used. RFG Frequency controller f* m* f act Drive 1 frequency controlled Frequency controller RFG m* f act Load equalization controller f* Drive 2 frequency controlled with higher-level load equalization control Block diagram of a load equalization control function 200 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks The frequency setpoint, torque setpoint and the control commands from the first converter are transferred to all of the other drive converters, for example, through a peer-to-peer coupling. In this case, the drive converters require an SCB2 board. f* SIMOVERT VC SCB 2 Drive 1 frequency controlled IM RS485 SIMOVERT VC SCB 2 Drive 2 frequency controlled IM Example of a load equalization control function with peer-to-peer connection (RS485) Siemens AG SIMOVERT MASTERDRIVES - Application Manual 201 3 Various special drive tasks 09.99 Master/slave drive SIMOVERT VC drive converters are required to implement master/slave drives. Each drive requires a speed tachometer (preferably a pulse encoder due to the enhanced accuracy). For master/slave drives, the load is uniformly distributed over the complete speed range by transferring the torque setpoint of the speed-controlled master drive to the torque-controlled slave drive. Speed controller n* m* n ist Master drive speed controlled m* Slave drive 1 torque controlled Block diagram of a master/slave drive In order that the slave drives don't accelerate up to excessive speeds when the load is removed (e. g. this could happen if the material web breaks), the closed-loop torque control of the slave drive can be implemented using a speed control, overdriven with n* with an appropriate torque limit. In this case, each slave drive requires, in addition to the torque setpoint of the master drive, the speed setpoint of the master drive. 202 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks From the master drive n* m* n * Speed controller m* n act Block diagram of a slave drive with over-driven closed-loop speed control The torque setpoint, speed setpoint and the control commands can be transferred to the slave drives, for example, via a peer-to-peer connection. n* SIMOVERT VC SCB 1/2 Master drive speed controlled IM Fiber optic / RS485 SIMOVERT VC SCB 1/2 IM Slave drive 1 torque-controlled Example of a master/slave drive with peer-to-peer connection (SCB1: Fiber-optic cable, SCB2: RS485) Siemens AG SIMOVERT MASTERDRIVES - Application Manual 203 3 Various special drive tasks 3.5.2 09.99 Group drive for a traction unit The four wheels of a traction unit are each driven by a motor. The two motors on one side of the traction unit are fed from one converter. As result of the different wheel loads, the wheels are subject to different wear factors. The load distribution of the motors on one side of the traction unit at maximum different wheel diameters and maximum velocity are to be determined. SIMOVERT Motor 1 IM i Motor 2 IM i v D max D min Block diagram of a traction unit side Drive data Rated motor output (8-pole) Pn = 40 kW Motor rated speed nn = 740 RPM Max. velocity vmax = 49.73 m/min Wheel diameter, new Dmax = 1.2 m Wheel diameter, after wear Dmin = 1.19 m Gearbox ratio i = 55.345 204 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks With v = vmax = 49.73/60 = 0.829 m/s, the motor speeds are given by: n1 = vmax 60 i 0.829 60 55.345 = 730 RPM = . Dmax 12 n2 = vmax 60 i 0.829 60 55.345 = 736 RPM = . Dmin 119 The following assumption is made to calculate the motor torques: n1 = n0 - nslip n M1 Mn n2 = n0 - nslip n M2 Mn M1 + M 2 = M load With nslip n = n0 n - nn = 750 - 740 = 10 RPM Thus, the motor torques are given by: M1 = M load (n2 - n1 ) M n + 2 2 nslip n M2 = M load (n2 - n1 ) M n - 2 2 nslip n With the differential torque: M = (n2 - n1 ) M n (736 - 730) M n = = 0.3 M n 2 nslip n 2 10 The motor with the larger wheel diameter is loaded with 30% higher rated motor torque and the motor with the lower wheel diameter, with 30% less rated motor torque. When low velocities are approached, the difference between the torques becomes less. The different loading must be taken into account when dimensioning the motors and defining the closed-loop control type. A vector control can no longer be optimally set for such different torque loads. If vector control is required, e. g. as a angular synchronous control is being used for both traction unit sides, then a configuration with 4 individual drives could be considered. The two single-motor drives on a traction unit side are designed as master/slave drives, and the master drives coupled through the angular synchronous control function. Siemens AG SIMOVERT MASTERDRIVES - Application Manual 205 3 Various special drive tasks 09.99 n / RPM 743 n1 n2 736 730 M1 =1.3 Mn M2 = 0.7 Mn M M load = 2 Mn Example with Mload = 2 Mn For the above example with Mload = 2 Mn, n0 is obtained as: n0 = n1 + nslip n ( 206 M load M + ) = 730 + 10 (1 + 0.3) = 743 RPM Mn 2 Mn Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks 3.5.3 Drive for an extraction tower with 10 mechanically-coupled motors An extraction tower is to be driven through a ring gear, pinion and gearbox with 10 motors, each 37 kW/ 6-pole, nn = 980 RPM. For redundancy reasons, each motor is provided with its own drive converter. In order to reduce gear stressing, the motors may only generate torque in one direction (motoring). Braking is not required. As result of the symmetrical configuration, regenerative operation of the individual motors can practically only occur under no-load conditions without material, as even the smallest differences here influence the stator frequency. The rated slip frequency of the motors is given by: f slip n = 1000 - 980 50 Hz = 1 Hz 1000 Thus, for a 0.01 Hz resolution stator frequency reference, the torque error lies below 1%. This can be achieved with a digital setpoint input. A peer-to-peer connection cannot be considered due to the required redundancy. Either a SINEC L2 DP bus connection via the CB1 board or an USS bus connection via the SCB2 board can be used. A V/f characteristic without slip compensation should be used. The ramp-up and ramp-down times must be the same for all drive converters. From the master Pinion SIMOVERT VC CB1 IM Drive 1 Rack SINEC L2-DP SIMOVERT VC CB1 IM Drive 2 Pinion to the other drives Implementing a bus connection via SINEC L2-DP Siemens AG SIMOVERT MASTERDRIVES - Application Manual 207 3 Various special drive tasks 3.6 09.99 Crank drive Mass m is to be raised by 2 r, within 0.25 s, to the upper deadpoint ( = ) using a crank drive configuration, starting from the lower deadpoint ( = 0). The return stroke is realized after a 0.25 s delay time. The stroke cycle time is 5 s. m l r , Drive data Mass to be moved m = 300 kg Crank radius r = 0.06 m Connecting rod length l = 0.3 m Positioning time ttotal = 0.25 s Cycle time T =5s 208 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks Angular velocity and angular acceleration Positioning should be as fast as possible, i. e. a triangular characteristic with tb = tv = 0.125 s is used. Upwards Interval Downwards ~ ~ max 0.125 s 0.25 s 0.5 s 0.625 s 0.75 s 5s t ~ ~ max t - max The maximum angular velocity during the upwards motion occurs at t = ttotal/2 = 0.125 s or at an angle of = /2. max = _ 2 t total 2 t ttotal Siemens AG SIMOVERT MASTERDRIVES - Application Manual 209 3 Various special drive tasks 09.99 Using the following equation = ttotal 2 = 2 2 max the maximum angular velocity is given by: max = 2 2 = = 2513 . s -1 ttotal 0.25 With max = max ttotal 2 the maximum angular acceleration is given by: max = max 2513 . = = 20106 . s -2 ttotal . 0125 2 The maximum speed at = max is given by: nmax = max 60 2513 . 60 = = 240 RPM 2 2 Due to the short accelerating- and decelerating times, precise positioning is only possible using a servo drive. 210 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks Converting the circular motion into a linear movement x, vx , a x xmin = l - r xmax = l + r x (lower deadpoint) (upper deadpoint) l r r cos r sin , From the diagram, x is obtained as follows: x = l 2 - r 2 sin 2 - r cos r2 l- sin 2 - r cos 2l (approximation) Thus, the following is obtained for velocity vx and acceleration ax: vx = r dx d r (sin - sin 2 ) 2l d dt ax = d v x d r2 r ( 2 cos + sin ) - ( 2 cos 2 + sin 2 ) d dt l 2 To evaluate these equations for vx and ax, angular velocity , angular acceleration and angle should be determined as a function of time t. To realize this, motion is sub-divided into four ranges. Siemens AG SIMOVERT MASTERDRIVES - Application Manual 211 3 Various special drive tasks Range 1 09.99 ) 2 upwards motion, accelerating ) 2 upwards motion, decelerating (0 t 0125 . s and 0 1 = max 1 = max t 1 1 = max t 2 2 Range 2 (0125 . s t 0.25 s and 2 = - max 2 = max (0.25 s - t ) 1 2 = - max (0.25 s - t ) 2 2 Range 3 (0.5 s 0.625 s and 3 ) 2 downwards motion, accelerating 3 = max 3 = max (t - 0.5 s) 1 3 = + max (t - 0.5 s) 2 2 Range 4 (0.625 s 0.75 s and 3 2 ) 2 downwards motion, decelerating 4 = - max 4 = max (0.75 s - t ) 1 4 = 2 - max (0.75 s - t ) 2 2 212 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks Calculating the load torque ax F =- ++= l r sin = r sin = l sin FT r FS + , With linear acceleration ax, the following force acts on mass m: F = m g + m ax linear force Using the angular relationships, which can be derived from the diagram above, the connecting rod force, the tangential force and the load force are obtained as follows: FS = F cos connecting rod force FT = FS sin( + ) = FS sin( - + ) M load = FT r = m ( g + a x ) sin( - + ) r cos tangential force load torque With r = arcsin ( sin ) l Siemens AG SIMOVERT MASTERDRIVES - Application Manual 213 3 Various special drive tasks 09.99 As result of the irregular angular velocity and the complex angular relationships, it is relatively complex to calculate the load torque. In order to evaluate the equations, it is recommended that a program such as Excel is used. If l is much greater than r (approx. from r / l < 0.1), the following approximation can be used as a function of to calculate the load torque: M load ( ) m r ( g sin + max r sin 2 + max r sin 2 ) for 0 M load ( ) m r ( g sin + max r ( - ) sin 2 - max r sin 2 ) for 2 2 The range 2 is a mirror image of range 0 . The maximum load torque can be determined using this approximation. The maximum lies within the range /4 < < /2. If the approximation is used for this particular case, the maximum for Mload is obtained at = 1.185 to 530 Nm. Angle , at which the maximum occurs, can also be determined by trial and error or by iteration. Without this approximation, for the particular case, the following load torque characteristic is obtained M Load / Nm Upwards motion Interval Downwards motion 600 400 200 0 0.125 0.25 0.375 0.5 0.625 0.75 t/ s -200 -400 -600 214 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks The maximum load torque when accelerating upwards is 572 Nm and when decelerating downwards, -572 Nm. Motor selection Selected motor: 1FK6 103-8AF71 with gearbox i=10 ( =0.95), nn=3000 RPM, Jmotor=0.01215 kgm2 Mn(100)=17 Nm, M0(100)=36 Nm, I0(100)=23.8 A, motor=0.94, JGearbox=0.00174 kgm2 Maximum motor speed: nmotor max = i nmax = 10 240 = 2400 RPM Calculating the motor torques Accelerating- and decelerating torques for the motor and the gearbox: M b motor + gearbox = M v motor + gearbox = J motor + gearbox i max = ( 0.01215 + 0.00174) 10 20106 . = 27.93 Nm Motor torque when accelerating: M motor ( t ) = Mb motor + M load ( t ) for 0 t 0125 . s and 0.5 s t 0.625 s 1) i 1) If the load torque is negative in these ranges, then at the appropriate locations, factor 1/ must be changed to . Motor torque when decelerating: M motor ( t ) = - M v motor + M load ( t ) i for 0125 . s t 0.25 s and 0.625 s t 0.75 s 2) 2) If the load torque is positive in these ranges, at the appropriate locations, factor must be changed into 1/. Siemens AG SIMOVERT MASTERDRIVES - Application Manual 215 3 Various special drive tasks 09.99 The highest motor torque is required when accelerating upwards at nmotor=2220 RPM: M motor max b = 27.93 + 572 = 881 . Nm 10 0.95 The highest motor torque in regenerative operation is required when decelerating during downwards motion: M motor max v = -27.93 + - 572 0.95 = -82.3 Nm 10 Frictional forces and forces due to other moving masses have been neglected. The selected motor is adequately dimensioned, as at nmotor=2220 RPM and 400 V supply voltage, it can be overloaded up to approx. 102 Nm. Motor torque characteristic for this particular case Mmotor / N m Upwards motion Interval Downwards motion 100 80 60 40 20 0 0.125 0.25 0.375 0.5 0.625 0.75 t / s -20 -40 -60 -80 -100 216 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks Calculating the maximum motor output The motor output is obtained from the motor torque and the angular velocity: Pmotor ( t ) = M motor ( t ) i ( t ) From the calculation, the maximum motor output is 21.2 kW when accelerating during upwards motion. The maximum motor output in regenerative operation is -19.8 kW. It occurs when decelerating during downwards motion. Motor output characteristic for this particular case P motor / k W Interval Upwards motion Downwards motion 25 20 15 10 5 0 0.125 0.25 0.375 0.5 -5 0.625 0.75 t/s -10 Regenerative operation -15 -20 -25 Thermally checking the motor The RMS torque for 1FT6 motors is generally obtained as follows for any torque characteristics: M RMS = M 2 dt T Siemens AG SIMOVERT MASTERDRIVES - Application Manual 217 3 Various special drive tasks 09.99 The integral expression can be approximately solved as follows by sub-dividing it into m segments: i =m 2 M dt i =1 M i2-1 + M i2 (ti - ti -1 ) 2 To evaluate this approximation, Mmotor must be available in tabular form. The following is obtained (refer to the table later on in the text): t = 0.25 s M t = 0.75 s 2 M dt + t =0 s 2 dt 1009 N 2 m2 s t = 0.5 s Thus, the following is obtained: 1009 = 14.2 Nm 5 M RMS The average motor speed is given by: nmot max naverage = 2 4 tb T = 2400 2 0125 . = 120 RPM 5 The calculated RMS torque lies, at naverage, below the MS1 characteristic. Thus, operation is permissible. Dimensioning the brake resistor Maximum braking power for the brake resistor: Pbr W max = Pbr motor max motor = 19.8 0.94 = 18.6 kW Using the motor output, the braking energy for a cycle can be calculated as follows: Wbr = t = 0.25 s t = 0.75 s t = 0.125 s motor t = 0.625 s Pmotor motor dt + P motor dt The integral expansion can be solved as follows by sub-dividing into m segments: i=m Pmotor i -1 + Pmotor i i =1 2 Pmotor motor dt motor 218 ( ti - ti - 1 ) Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks The calculation results in (refer to the table later in the text): Wbr 14 . kWs The following must be true for the brake resistor: Wbr 14 . = = 0.28 kW Pbr cont . 5 T With P20 36 Pbr cont . = (with an internal brake resistor) the following is obtained 36 0.28 = 10.08 kW P20 Further, the following must also be true: Pbr W max = 18.6 kW 15 . P20 Thus, a braking unit is selected with P 20 = 20 kW (6SE7023-3EA87-2DA0). The internal brake resistor is adequate. Selecting the drive converter The highest required motor torque is given by: M motor max M 0 (100) = 881 . = 2.45 36 The following value is obtained from the torque/current characteristic: I motor max I 0 (100) = 2.7 and I motor max = I 0 (100) 2.7 = 238 . 2.7 = 64.3 A Siemens AG SIMOVERT MASTERDRIVES - Application Manual 219 3 Various special drive tasks 09.99 Drive converter selected: 6SE7024-7ED51 PV n=22 kW; IV n=47 A; IV max=75.2 A Table for the approximate computation of the integral expression t 220 Mload Mmotor Pmotor M 2 dt P dt 0 0.00 27.93 0.00 0.00 0.00 0.025 10.52 29.03 1.46 20.29 0.00 0.05 61.48 34.40 3.46 45.62 0.00 0.072 180.63 46.94 6.80 82.87 0.00 0.094 394.58 69.46 13.13 160.19 0.00 0.116 572.09 88.15 20.56 298.73 0.00 0.1205 564.89 87.39 21.17 333.39 0.00 0.125 530.16 83.73 21.04 366.35 0.00 0.125 95.87 -17.84 -4.48 366.35 0.00 0.1375 -115.74 -38.92 -8.80 377.81 -0.08 0.15 -208.92 -47.77 -9.61 401.54 -0.19 0.15625 -210.25 -47.90 -9.03 415.85 -0.24 0.18125 -79.76 -35.50 -4.91 460.29 -0.40 0.20625 6.72 -27.22 -2.39 485.31 -0.49 0.25 0.00 -27.93 0.00 518.57 -0.54 0.5 0.00 27.93 0.00 518.57 -0.54 0.54375 -6.72 27.29 2.40 551.93 -0.54 0.56875 79.76 36.32 5.02 577.73 -0.54 0.59375 210.25 50.06 9.44 625.54 -0.54 0.6 208.92 49.92 10.04 641.16 -0.54 0.6125 115.74 40.11 9.073 666.79 -0.54 0.625 -95.87 18.82 4.73 679.06 -0.54 0.625 -530.16 -78.29 -19.68 679.06 -0.54 0.6295 -564.89 -81.59 -19.77 707.83 -0.62 0.634 -572.09 -82.28 -19.19 738.04 -0.71 0.656 -394.58 -65.41 -12.36 859.57 -1.03 0.678 -180.63 -45.09 -6.53 929.00 -1.23 0.7 -61.48 -33.77 -3.39 963.90 -1.33 0.725 -10.52 -28.93 -1.45 988.62 -1.39 0.75 0.00 -27.93 0.00 1008.83 -1.40 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3.7 3 Various special drive tasks Rotary table drive This involves a so-called applicator drive with a maximum of 36000 applications per hour. An application means a rotation through 18 degrees within 50 ms and a subsequent interval of 50 ms. Rotary table (application roll) Gearbox Motor Drive data Angle for an application (i.e. attaching a label) = 18 degrees Rotary table moment of inertia Jload = 0.05 kgm2 Positioning time ttotal = 50 ms Cycle time T = 100 ms Siemens AG SIMOVERT MASTERDRIVES - Application Manual 221 3 Various special drive tasks 09.99 Positioning must be as fast as possible, i. e., a triangular characteristic with tb=tv=ttotal/2=25 ms is used. load load max tb t total tv t T load load max t - load max With the following angle = load max ttotal = 2 10 the maximum angular velocity is given by: load max 2 2 10 = = = 12.57 s -1 ttotal 0.05 With load max = load max ttotal 2 the following is obtained for the maximum angular acceleration: load max = 222 2 load max ttotal = 2 12.57 = 502.7 s -2 0.05 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks The maximum load speed at = load max is obtained as follows: nload max = load max 60 = 120 RPM 2 As a result of the short accelerating- and decelerating times, precise positioning is only possible using a servo drive. Calculating the load torques when accelerating and decelerating As acceleration is the same as deceleration, the load torques are given by: Mb load = Mv load = J load load max = 0.05 502.7 = 2513 . Nm Selecting the motor The following motor is selected (refer to Calculating the motor torques): 1FT6 062-6AC7 with gearbox i=5 (=0.95), nn=2000 RPM, Jmotor=0.00085 kgm2 Jgearbox=0.0003 kgm2, Mn(100)=5.2 Nm, In(100)=2.6 A, motor=0.84 The gearbox ratio was selected close to the optimum: iopt = J load = J motor + J gearbox 0.05 = 6.6 0.00085 + 0.0003 Thus, the maximum motor speed is given by: nmotor max = i nload max = 5 120 = 600 RPM Calculating the motor torques when accelerating and decelerating Accelerating- and decelerating torques for the motor: M b motor = M v motor = J motor i load max = 0.00085 5 502.7 = 214 . Nm Siemens AG SIMOVERT MASTERDRIVES - Application Manual 223 3 Various special drive tasks 09.99 Accelerating- and decelerating torque for the gearbox: M b gearbox = M v gearbox = J gearbox i load max = 0.0003 5 502.7 = 0.754 Nm Motor torque when accelerating: Mmotor = M b motor + M b gearbox + M b load = 214 . + 0.754 + 2513 . 1 i 1 = 818 . Nm 5 0.95 Motor torque when decelerating: Mmotor = - M v motor - M v gearbox - M v load = -214 . - 0.754 - 2513 . i 0.95 = -7.67 Nm 5 (regenerative operation) The highest motor torque is required when accelerating. The selected motor is adequately dimensioned, as it can be overloaded up to approx. 24 Nm at nmax=600 RPM and a 400 V supply voltage. Selecting the drive converter A motor current, for the highest required motor torque, is obtained as follows: I motor max = I motor n M motor max M motor n = 2.6 818 . = 41 . A 5.2 (for this motor there is hardly any saturation effect) Selected drive converter: 6SE7013-0EP50 (Compact Plus) PV n=1.1 kW; IV n=3 A, IV max=4.8 A (160% overload capability) 224 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks Dimensioning the brake resistor Max. braking power for the brake resistor during deceleration from load max to 0: Pbr W max = Pbr motor max motor = = Mmotor v nmotor max 9550 motor 7.67 600 0.84 = 0.405 kW 9550 Braking diagram P br W 0.405 kW tb t tv T Braking energy for a cycle (corresponds to the area in the braking diagram): Wbr = = 1 Pbr W max tv 2 1 0.405 0.025 = 0.005 kWs 2 The following must be true for the brake resistor: Wbr 0.005 = = 0.05 kW Pbr cont . T 01 . Siemens AG SIMOVERT MASTERDRIVES - Application Manual 225 3 Various special drive tasks 09.99 With Pbr cont. = P20 4.5 (for an external brake resistor) the following is obtained 4.5 0.05 = 0.225 kW P20 Thus, the smallest braking resistor is selected with P 20 = 5 kW (6SE7018-0ES87-2DC0). Thermally checking the motor Torque characteristic M motor 8.18 Nm 0.025 s t 0.025 s -7.67 Nm 0.1 s The RMS torque and the average motor speed is given by: M RMS = 818 . 2 0.025 + 7.67 2 0.025 = 5.61 Nm 01 . nmot max naverage = 2 2 tb T = 600 0,025 = 150 RPM 01 . The calculated RMS torque lies, at naverage, below the MS1 characteristic. Thus, operation is thermally permissible. 226 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3.8 3 Various special drive tasks Pivot drive This involves a pivot drive with an 180 degree rotational movement forwards and backwards through a horizontal axis. A servo drive is to be used. After the drive has rotated, it is held in the initial position by a mechanical brake. max = 180 degrees m2 m2 g r2 = 650 Fz m1 r1 = 650 m1 g Initial position of the pivot drive Drive data Angle of rotation max = 180 degrees Weight m1 = 170 kg Opposing weight m2 = 130 kg Supplementary force Fz = 500 N Pivot time when moving through 180 degrees ttotal =3s Delay time before the start (drive on) tw =3s Interval between forwards and return motion (drive on) tp =1s Cycle time T = 45 s Siemens AG SIMOVERT MASTERDRIVES - Application Manual 227 3 Various special drive tasks 09.99 Positioning should be as fast as possible, i. e., a triangular accelerating characteristic with tb = tv = t total/2 = 1.5 s is used. Thus, the following motion diagram is obtained. load Delay time Forwards Interval Backwards max load ~ ~ t total tb tw - max load tv t total t tp Drive on Brake on T load ~ ~ max load t - max load The maximum angular velocity and the maximum angular acceleration are given by: load max = 2 max 2 = = 2.094 s -1 3 ttotal load max = load max 2 2.094 2 = = 1396 . s -2 ttotal 3 228 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks The maximum load speed at load = load max is given by: nload max = loadt max 60 2.094 60 = = 20 RPM 2 2 As the geometrical dimensions of the weights are not known, it is assumed that their centers of gravity act at single points. Thus, the load moment of inertia is calculated as follows: J load = m1 r12 + m2 r22 = 170 0.652 + 130 0.652 = 126.75 kgm2 Calculating the load torques Fz r1 m1 g r2 m2 g M load = ( m1 g + Fz ) r1 cos - m2 g r2 cos = ((170 9.81 + 500) 0.65 - 130 9.81 0.65) cos = 580.06 Nm cos The accelerating- and decelerating torque of the load is given by: Mb load = M v load = J load load max = 126.75 1396 . = 177 Nm Siemens AG SIMOVERT MASTERDRIVES - Application Manual 229 3 Various special drive tasks 09.99 Selecting the motor The following motor is selected (refer to Calculating the motor torques): 1FT6 062-1AF71 with gearbox i=150 ( =0.95); nn=3000 RPM; Jmotor+brake=0.00105 kgm2 Mn(100)=4.6 Nm; In(100)=3.4 A; motor=0.88 Maximum motor speed: nmotor max = i nload max = 150 20 = 3000 RPM Calculating the motor torques Accelerating- and decelerating torque for the motor: Mb motor = M v motor = Jmotor i load max = 0.00105 150 1396 . = 0.22 Nm Holding torque for the motor during delay time at = 0 and during the no-load interval at =180 degrees: M H motor = M load 580.06 = = 387 . Nm i 150 Motor torque when accelerating forwards: M motor ( t ) = Mb motor + ( M load ( t ) + Mb load ) = 0.22 + (580.06 cos + 177) 1 i for 3 s t 4.5 s and 0 2 1 150 0.95 = 146 . Nm + 4.07 Nm cos With = 230 1 load max (t - t w ) 2 2 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks Motor torque when decelerating forwards: M motor ( t ) = - M v motor + ( M load ( t ) - M v load ) = -0.22 + (580.06 cos - 177) i for 4.5 s t 6 s and 2 0.95 150 = -134 . Nm + 3.67 Nm cos (regenerative operation) With 1 = - load max (t w + ttotal - t ) 2 2 Motor torque when accelerating backwards: M motor ( t ) = - Mb motor + ( M load ( t ) - Mb load ) = -0.22 + (580.06 cos - 177) 1 i for 7 s t 8.5 s and 2 1 150 0.95 = -146 . Nm + 4.07 Nm cos With 1 = - load max (t - ttotal - t p - t w ) 2 2 Motor torque when decelerating backwards: M motor ( t ) = M v motor + ( M load ( t ) + M v load ) = 0.22 + (580.06 cos + 177) = 134 . Nm + 3.67 Nm cos i for 8.5 s t 10 s and 0 2 0.95 150 (regenerative operation) With 1 = load max ( 2 t total + t p + t w - t ) 2 2 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 231 3 Various special drive tasks 09.99 The highest motor torque is required when accelerating forwards at = 0 and when accelerating backwards at = : M motor max b = 146 . + 4.07 cos 0 = 553 . Nm The highest motor torque in regenerative operation is required when decelerating forwards at = and when decelerating backwards at = 0: M motor max v = -134 . + 3.67 cos = -5.01 Nm The gearbox moment of inertia was neglected. Motor torque characteristic for this particular case Delay time Forwards Interval Backwards M motor / Nm 6 4 2 ~ ~ 0 2 4 6 8 10 12 14 45 t/s -2 -4 -6 232 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks Selecting the drive converter For the highest required motor torque, the motor current is given by: I motor max = I motor n M motor max M motor n = 3.4 553 . = 4.09 A 4.6 (for this motor there is hardly any saturation effect) Drive converter selected: 6SE7013-0EP50 (Compact Plus) PV n=1.1 kW; IV n=3 A; IV max=4.8 A (160% overload capability) Dimensioning the brake resistor The motor output must be determined when dimensioning the brake resistor. Pmotor = M motor load i With load = 0 for 0 t 3 s and = 0 load = load max (t - t w ) for 3 s t 4.5 s and 0 load = load max (t w + ttotal - t ) for 4.5 s t 6 s and load = 0 for 6 s t 7 s and = load = - load max (t - ttotal - t p - t w ) for 7 s t 8.5 s and load = - load max (2 ttotal + t p + t w - t ) for 8.5 s t 10 s and Siemens AG SIMOVERT MASTERDRIVES - Application Manual 2 2 2 0 2 233 3 Various special drive tasks 09.99 Motor output characteristic for this particular case Delay time Forwards Interval Backwards Pmotor / W 1000 800 600 400 ~ ~ 200 0 2 4 6 8 10 12 14 45 t/s -200 -400 -600 -800 -1000 Regenerative operation Regenerative operation occurs when decelerating in the ranges 4.5 s t 6 s and 8.5 s t 10 s . From the calculation, the maximum motor output in regenerative operation is -875 W. Thus, the maximum braking power for the brake resistor is given by: Pbr W max = Pbr motor max motor = -875 0.88 = -770 W The braking energy for a cycle can be calculated as follows from the motor output: t =6 s Wbr = Pmotor motor dt + t = 4 .5 s t = 10 s P motor motor dt t = 8.5 s The integrals can be approximately calculated as follows by sub-dividing into m segments: i=m Pmotor i -1 + Pmotor i i =1 2 Pmotor motor dt motor ( ti - ti - 1 ) To evaluate this approximation formula, Pmotor must be available in tabular form. 234 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks The calculation provides the following results (refer to the table later in the text): Wbr -757.67 - 757.67 -1515.34 Ws The following must be true for the brake resistor: Wbr 1515.34 = = 33.67 W Pbr cont . 45 T With Pbr cont. = P20 4.5 (with an external brake resistor) the following is obtained 4.5 33.67 = 1515 . W P20 Further, the following must be true: Pbr W max = 770 W 15 . P20 Thus, the smallest braking resistor is selected with P20 = 5 kW (6SE7018-0ES87-2DC0). Thermally checking the motor For the RMS torque for 1FT6 motors, generally the following torque characteristics are valid: M RMS = M 2 dt T The integral expression can be approximately calculated by sub-dividing into m segments: i=m 2 M dt i =1 Mi2-1 + Mi2 ( ti - t i - 1 ) 2 To evaluate this approximation formula, Mmotor must be available in tabular form. The calculation provides the following results (refer to the table later in the text): t = 10 s M 2 dt 18354 . N 2 m2 s t =0 s Siemens AG SIMOVERT MASTERDRIVES - Application Manual 235 3 Various special drive tasks 09.99 Thus, the following is obtained: M RMS 18354 . 2.02 Nm 45 The calculated RMS torque is less than the torque permissible in S1 duty. Thus, operation is permissible. Table for approximately calculating the integral expressions t Mmotor Pmotor M 2 dt P dt 0.00 3.87 0.00 0.00 0.00 3.00 3.87 0.00 44.86 0.00 3.00 5.53 0.00 44.86 0.00 3.15 5.53 173.79 49.45 0.00 3.30 5.52 347.11 54.03 0.00 3.45 5.49 517.59 58.55 0.00 3.60 5.40 679.16 62.94 0.00 3.75 5.22 820.36 67.03 0.00 3.90 4.90 923.40 70.63 0.00 4.05 4.39 964.32 73.51 0.00 4.20 3.64 915.58 75.50 0.00 4.35 2.66 751.75 76.56 0.00 4.50 1.46 459.25 76.88 0.00 4.50 -1.34 -421.21 76.88 0.00 4.65 -2.42 -684.52 77.76 -72.98 4.80 -3.31 -831.70 79.41 -173.05 4.95 -3.98 -875.02 81.78 -285.69 5.10 -4.44 -837.41 84.74 -398.71 5.25 -4.73 -743.75 88.10 -503.07 5.40 -4.90 -615.63 91.70 -592.79 5.55 -4.98 -469.15 95.42 -664.38 5.70 -5.01 -314.61 99.18 -716.11 5.85 -5.01 -157.52 102.95 -747.27 6.00 -5.01 0.00 106.72 -757.67 Continued on the next page 236 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks t Mmotor Pmotor M 2 dt P dt 6.00 -3.87 0.00 106.72 -757.67 7.00 -3.87 0.00 121.68 -757.67 7.00 -5.53 0.00 121.68 -757.67 7.15 -5.53 173.79 126.27 -757.67 7.30 -5.52 347.11 130.85 -757.67 7.45 -5.49 517.59 135.37 -757.67 7.60 -5.40 679.16 139.75 -757.67 7.75 -5.22 820.36 143.84 -757.67 7.90 -4.90 923.40 147.44 -757.67 8.05 -4.39 964.32 150.33 -757.67 8.20 -3.64 915.58 152.32 -757.67 8.35 -2.66 751.75 153.38 -757.67 8.50 -1.46 459.25 153.70 -757.67 8.50 1.34 -421.21 153.70 -757.67 8.65 2.42 -684.52 154.58 -830.65 8.80 3.31 -831.70 156.22 -930.72 8.95 3.98 -875.02 158.60 -1043.36 9.10 4.44 -837.41 161.56 -1156.38 9.25 4.73 -743.75 164.92 -1260.74 9.40 4.90 -615.63 168.52 -1350.46 9.55 4.98 -469.15 172.24 -1422.05 9.70 5.01 -314.61 176.00 -1473.78 9.85 5.01 -157.52 179.77 -1504.94 10.00 5.01 0.00 183.54 -1515.34 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 237 3 Various special drive tasks 3.9 09.99 Spindle drive with leadscrew This drive involves a plastic machining unit, with vertical- and horizontal spindle drives without gearbox. Servo drives with Compact Plus as 2-axis system should be used. Motor for the vertical drive Motor for the horizontal drive Principle of operation Vertical drive Drive data Total mass to be raised mtotal = 116 kg Spindle diameter D = 0.04 m Spindle pitch h = 0.01 m Spindle length l = 1.2 m Spindle efficiency Sp = 0.9 Moment of inertia of the coupling (acc. to the manufacturer) JK = 0.00039 kgm2 Max. elevating vmax = 0.33 m/s Accelerating time tb = 0.1 s Decelerating time tv = 0.1 s Max. distance raise hmax = 0.651 m No-load interval between elevating and lowering tP = 10 s Cycle time T = 20 s 238 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks Elevating force, acceleration force: FH = mtotal g = 116 9.81 = 1138 N Fb = Fv = mtotal a = mtotal v max 0.33 = 116 = 382.8 N tb 0,1 Spindle pitch angle: SW = arctan h 0.01 = arctan = 0.07941 rad D 0.04 Friction angle of the spindle: = arctan h 0.01 - SW = arctan - 0.07941 = 0.00416 rad D Sp 0.04 0.95 Calculating the torques for spindle and coupling when accelerating and decelerating Angular velocity of the spindle at vmax: max Sp = 2 vmax 2 0.33 = = 207.3 s -1 h 0.01 Angular acceleration of the spindle: b sp = max sp 207.3 = = 2073 s -2 tb 0.1 According to the manufacturer, the spindle moment of inertia 40x10 is as follows: J sp = 119.27 10-6 + 1.607 10-3 l = 119.27 10-6 + 1.607 10-3 1.2 = 0.00205 kgm2 The first component is a constant value and the second component represents the formula for the moment of inertia of the solid cylinder. Accelerating- and decelerating torque for spindle+coupling: Mb sp + c = M v sp + c = ( J sp + Jc ) b sp = (0.00205 + 0.00039) 2073 = 5.06 Nm Siemens AG SIMOVERT MASTERDRIVES - Application Manual 239 3 Various special drive tasks 09.99 Motor selection The following motor is selected (refer to Calculating the motor torques): 1FK6 042-6AF71, nn=3000 RPM, Mn(100)=2.6 Nm, M0(100)=3.2 Nm, I0(100)=2.7 A, motor=0.89 Jmotor+brake=0.000368 kgm2, Mmotor perm.=10.3 Nm Motor speed at vmax: nmotor max = max sp 60 207.3 60 = = 1980 RPM 2 2 Calculating the motor torques during the constant-velocity phase Upwards: M motor = FH D 0.04 tan( SW + ) = 1138 tan( 0.07941 + 0.00416) = 191 . Nm 2 2 Downwards: M motor = FH D 0.04 tan( SW - ) = 1138 tan( 0.07941 - 0.00416) = 172 . Nm 2 2 (reg. op.) Calculating the motor torques when accelerating and decelerating Motor accelerating- and decelerating torque: Mb motor = M v motor = Jmotor b sp = 0.000368 2073 = 0.763 Nm Upwards, motor torque when accelerating: M motor = M b motor + M b sp + c + ( Fb + FH ) = 0.763 + 5.06 + (382.8 + 1138) 240 D tan( SW + ) 2 0.04 tan( 0.07941 + 0.00416) = 8.37 Nm 2 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks Upwards, motor torque when decelerating: M motor = - M v motor - M v sp+ c + ( - Fv + FH )1) = -0.763 - 5.06 + ( -382.8 + 1138) D tan( SW + sign(... )) 2 0.04 tan( 0.07941 + 0.00416) = -4.56 Nm 2 (reg. op.) Downwards, motor torque when accelerating: M motor = - M b Motor - M b sp + c + ( - Fb + FH )1) = -0.763 - 5.06 + ( -382.8 + 1138) D tan( SW - sign(... )) 2 0.04 tan( 0.07941 - 0.00416) = -4.68 Nm 2 Downwards, motor torque when decelerating: M motor = M v Motor + M v sp+ c + ( Fv + FH ) = 0.763 + 5.06 + (382.8 + 1138) D tan( SW - ) 2 0.04 tan( 0.07941 - 0.00416) = 812 . Nm 2 1) If the expression in brackets has a negative sign, the sign of (reg. op.) changes The highest motor torque is required when accelerating upwards. The highest motor torque in regenerative operation is required for deceleration when moving downwards. The selected motor is adequately dimensioned, as it can be overloaded up to approx. 10.3 Nm at nmax=1980 RPM and for a 400 V line supply voltage. Selecting the drive converter The following is valid for the highest required motor torque: M motor max M 0 (100) = 8.37 = 2.62 3.2 Thus, the following value is obtained from the torque/current characteristic: I motor max I 0 (100) = 2.7 and Siemens AG SIMOVERT MASTERDRIVES - Application Manual I motor max = I 0 (100) 2.7 = 2.7 2.7 = 7.3 A 241 3 Various special drive tasks 09.99 Drive converter selected: 6SE7013-0EP50 (Compact Plus) PV n=1.1 kW; IV n=3 A; IV max=9 A (300% overload capability) Calculating of the braking power and the braking energy The brake resistor is used when the drive decelerates when descending or ascending. Velocity-time diagram v Elevating Interval Lowering vmax t total tk tv tb tk tv ~ ~ ~ ~ tb t tp t total - vmax T The traversing time is given by: ttotal = hmax + vmax tb 0.651 + 0.33 01 . = = 2.07 s vmax 0.33 Time for constant-velocity motion: t k = ttotal - 2 tb = 2.07 - 2 01 . = 187 . s Max. braking power for the brake resistor when decelerating from vmax to 0: Pbr W max (vmax 0) = Pbr motor max ( v max 0) motor = = 242 M motor v ( v max 0 ) nmotor max 9550 motor 4.56 1980 0.89 = 0.84 kW 9550 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks Max. braking power for the brake resistor when decelerating from -vmax to 0: Pbr W max (-vmax 0) = Pbr motor max ( - v max 0) motor = = M motor v ( - v max 0) nmotor max 9550 motor 812 . 1980 0.89 = 15 . kW 9550 Braking power while travelling at constant velocity downwards: Pbr W konst ( v =- vmax ) = Pbr Motor konst ( v =- vmax ) Motor = = M Motor v ( v =- vmax ) n Motor max 9550 Motor 172 . 1980 0.89 = 0.317 kW 9550 Braking diagram P Elevating br W Interval Lowering 1.5 kW 0.84 kW ~ ~ ~ ~ 0.317 kW tv tk tv T Braking energy for a cycle (corresponds to the area in the braking diagram): Wbr = = 1 1 Pbr W max ( vmax 0 ) t v + Pbr W konst ( v =- vmax ) t k + Pbr W max ( - vmax 0 ) t v 2 2 1 1 0.84 01 . + 0.317 187 . + 15 . 01 . = 0.71 kWs 2 2 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 243 3 Various special drive tasks 09.99 Thermally checking the motor Torque characteristic Elevating M motor Interval Lowering 8.37 Nm 8.12 Nm 2.07 s 1.91 Nm 1.87 s 0.1 s 1.87 s ~ ~ ~ ~ 0.1 s 1.72 Nm 10 s 0.1 s t 0.1 s -4,56 Nm -4.68 Nm 2.07 s 20 s The RMS torque is obtained from the torque characteristic as follows: M RMS = 8.37 2 01 . + 191 . 2 187 . + 4.56 2 01 . + 4.68 2 01 . + 172 . 2 187 . + 812 . 2 01 . = 123 . Nm 20 The average motor speed is given by: naverage = 2( nmotor max 2 t b 2 + nmotor max t k ) T = 2( 1980 01 . 2 + 1980 187 . ) 2 = 390 RPM 20 The calculated RMS torque lies, at naverage, below the MS1 characteristic. Thus, operation is permissible. 244 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks 12 M mot, M per. in Nm 10 8 6 4 2 0 0 500 1000 1500 2000 2500 3000 Motor speed in RPM Absolute motor torque and dynamic limiting characteristic 3,5 Motor torque in Nm 3 2,5 2 M per. S1 M RMS / n average 1,5 1 0,5 0 0 500 1000 1500 2000 2500 3000 Motor speed in RPM RMS motor torque at naverage and Mperm S1 characteristic Siemens AG SIMOVERT MASTERDRIVES - Application Manual 245 3 Various special drive tasks 09.99 Horizontal drive Drive data Mass to be moved mtotal = 100 kg Spindle diameter D = 0.032 m Spindle pitch h = 0.01 m Spindle length l =1m Spindle efficiency Sp = 0.95 Slide friction wF = 0,02 Coupling moment of inertia (acc. to the manufacturer) JK = 0.0002 kgm2 Max. traversing velocity vmax = 0.4 m/s Accelerating time tb = 0.16 s Decelerating time tv = 0.16 s Max. distance moved hmax = 0.427 m Interval between feed and retraction tP =5s Cycle time T = 20 s Frictional force, accelerating force: FR = mtotal g wF = 100 9.81 0.02 = 19.62 N Fb = Fv = mtotal a = mtotal v max 0.4 = 100 = 250 N tb . 016 Spindle pitch angle: SW = arctan h 0.01 = arctan = 0.09915 rad D 0.032 Friction angle of the spindle: = arctan h 0.01 - SW = arctan - 0.09915 = 0.00518 rad D Sp 0.032 0.95 Calculating the torques for spindle and coupling when accelerating and decelerating Angular velocity of the spindle at vmax: max sp = 246 2 vmax 2 0.4 = = 2513 . s -1 h 0.01 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks Angular acceleration of the spindle: b sp = max sp 2513 . = = 1571 s -2 016 . tb According to the manufacturer, the spindle moment of inertia 32x10 is calculated as follows: J sp = 33573 . 10-6 + 0.712 10-3 l = 33573 . 10-6 + 0.712 10-3 1 = 0.000746 kgm2 The first component is a constant value and the second component represents the formula for the moment of inertia of a solid cylinder. Accelerating torque and decelerating torque for spindle+coupling: M b sp + c = M v sp + c = ( J sp + J c ) b sp = (0.000746 + 0.0002) 1571 = 149 . Nm Selecting the motor The following motor is selected (refer to calculating the motor torques): 1FK6 042-6AF71, nn=3000 RPM, Mn(100)=2.6 Nm, M0(100)=3.2 Nm, I0(100)=2.7 A, motor=0.89 Jmotor+brake=0.000368 kgm2, Mmotor perm.=10.3 Nm Motor speed at vmax: nmotor max = max sp 60 2513 . 60 = = 2400 RPM 2 2 Calculating the motor torques during the constant-velocity phase M motor = FR D 0.032 tan( SW + ) = 19.62 tan( 0.09915 + 0.00518) = 0.0329 Nm 2 2 Calculating the motor torques when accelerating and decelerating Accelerating- and decelerating torque for the motor: Mb motor = M v motor = Jmotor b sp = 0.000368 1571 = 0.578 Nm Siemens AG SIMOVERT MASTERDRIVES - Application Manual 247 3 Various special drive tasks 09.99 Motor torque when accelerating: Mmotor = M b motor + M b sp+ c + ( Fb + FR ) = 0.578 + 149 . + ( 250 + 19.62) D tan( SW + ) 2 0.032 tan( 0.09915 + 0.00518) = 2.52 Nm 2 Motor torque when decelerating: Mmotor = - M v Motor - M v sp+ c + ( - Fv + FR )1) = -0.578 - 149 . + ( -250 + 19.62) D tan( SW + sign(... )) 2 0.032 tan( 0.09915 - 0.00518) = -2.42 Nm 2 1) If the expression in brackets has a negative sign, the sign of (reg. op.) changes The highest motor torque is required when accelerating. The highest motor torque in the regenerative mode is required when decelerating. The selected motor is adequately dimensioned, as it can be overloaded up to approx. 10.3 Nm for nmax=2400 RPM and a 400 V line supply voltage. Selecting the drive converter The highest required motor torque is given by: M motor max M 0 (100) = 2.52 = 0.788 32 . The following value is obtained from the torque/current characteristic: I motor max I 0 (100) = 0.788 and I motor max = I 0 (100) 0.788 = 2.7 0.788 = 213 . A Drive converter selected: 6SE7012-0TP50 (Compact Plus) PV n=0.75 kW; IV n=2 A; IV max=6 A (300% overload capability) 248 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks This inverter is connected to the previously selected drive converter for the horizontal drive (2-axis operation) Calculating of the braking power and the braking energy The brake resistor is used when the drive decelerates. Velocity-time diagram v Interval Forwards Backwards vmax t total tb tv ~ ~ tk tb tk tv t tp t total - vmax T The traversing time is obtained as follows: ttotal = smax + vmax tb 0.427 + 0.4 016 . = = 123 . s vmax 0.4 Time where the drive moves with a constant velocity: t k = ttotal - 2 tb = 123 . - 2 016 . = 0.91 s Max. braking power for the brake resistor when decelerating from vmax to 0: Pbr W max (vmax 0) = Pbr motor max ( v max 0) motor = = M motor v ( v max 0) nmotor max 9550 motor 2.42 2400 0.89 = 0.541 kW 9550 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 249 3 Various special drive tasks 09.99 Braking diagram P br W Forwards Interval Backwards ~ ~ ~ ~ 0.541 kW t tv tv T Braking energy for a cycle (corresponds to the area in the braking diagram): 1 Wbr = 2 Pbr W max (vmax 0) tv 2 1 = 2 0.541 016 . = 0.087 kWs 2 250 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks Thermally checking the motor Torque characteristic Forwards M Interval Backwards motor 2.52 Nm 2.42 Nm 1.23 s 0.16 s 0.91 s 0.16 s 0.91 s ~ ~ ~ ~ 0.0329 Nm 5s t -0.0329 Nm 0.16 s 0.16 s -2.42 Nm -2.52 Nm 1.23 s 20 s The RMS torque is obtained from the torque characteristic as follows: M RMS = 2 ( 2.32 2 016 . + 0.0329 2 0.91 + 2.42 2 016 . ) = 0.442 Nm 20 The average motor speed is given by: naverage = 2( nmotor max 2 t b 2 + nmotor max t k ) T = 2( 2400 016 . 2 + 2400 0.91) 2 = 257 RPM 20 The calculated RMS torque lies, at naverage, below the MS1 characteristic. Thus, operation is permissible. Siemens AG SIMOVERT MASTERDRIVES - Application Manual 251 3 Various special drive tasks 09.99 12 M mot, M per. in Nm 10 8 6 4 2 0 0 500 1000 1500 2000 2500 3000 Motor speed in RPM Absolute motor torque and dynamic limiting characteristic 3,5 Motor torque in Nm 3 2,5 2 M per. S1 M RPM / n average 1,5 1 0,5 0 0 500 1000 1500 2000 2500 3000 Motor speed in RPM RMS motor torque at naverage and Mperm S1 characteristic 252 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks Dimensioning the brake resistor When dimensioning the common brake resistor, it is assumed that the max. brake power of the two drives can occur simultaneously. Thus, the following is true: Pbr W max tot . = Pbr W max1 + Pbr W max 2 = 15 . + 0.541 = 2.041 kW 15 . P20 Further, the following must be true for the brake resistor: Wbr1 + Wbr 2 0.71 + 0.087 = = 0.0399 Pbr cont T 20 With Pbr cont = P20 4.5 (for Compact Plus, only an external brake resistor) the following is obtained 4.5 0.0399 = 018 . kW P20 Thus, the smallest brake resistor is selected with P 20 = 5 kW (6SE7018-0ES87-2DC0). Siemens AG SIMOVERT MASTERDRIVES - Application Manual 253 3 Various special drive tasks 3.10 09.99 Cross-cutter drive This drive involves a cross-cutter drive with a directly driven knife roll. A servo drive is to be used. load Knife roll D v web s Drive data Reel diameter D = 150 mm Reel weight mW = 1 kg Web velocity vweb = 100 m/min Cut length s = 250 mm The feed time for the 250 mm cut is calculated from the web velocity as follows: ts = s vweb = 250 mm 0.25 60 = = 015 . s 100 m / min 100 When cutting, the knife roll must rotate with the same circumferential velocity as the web velocity. If this velocity is maintained, then a feed time per cut is obtained as follows: t s = 254 150 mm 015 . 60 D = = = 0.283 s 100 m / min 100 v web Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks After each cut, the knife roll must be accelerated so that it has the same velocity as the web, and it must then be decelerated to a circumferential velocity which is the same as the feed velocity. This should be realized as fast as possible, e.g. a triangular characteristic is used with tb = t v = ts 150 ms - 10 ms = - 10 ms = 65 ms 2 2 The 10 ms takes into account deadtimes and controller stabilization. load Area load max load min Cut t t b t ^2 = v t s load load max t - load max The minimum angular velocity of the knife roll is obtained from the web velocity: load min = v web 2 2 100 2 = 100 m / min = = 22.22 s -1 150 mm 60 015 D . Siemens AG SIMOVERT MASTERDRIVES - Application Manual 255 3 Various special drive tasks 09.99 The maximum angular velocity is derived from the angle through which the knife roll has moved per revolution. This angle corresponds to 2. Thus, the following is true: 2 = load min t s + load max - load min 2 (t b + t v ) (corresponds to the area under load ) and load max = load min + 2 2 (2 - load min t s ) = 22.22 + (2 - 22.22 015 . ) = 67.6 s -1 013 . tb + tv The maximum angular acceleration of the knife roll is given by: load max = load max - load min tb = 67.6 - 22.22 = 698.2 s -2 0.065 The maximum speed of the knife roll is: nload max = load max 60 2 = 67.6 60 = 645.6 RPM 2 Calculating the load torques when acelerating and decelerating Moment of inertia of the load: J load = 1 D 1 015 . 2 mW ( ) 2 = 1 ( ) = 0.00281 kgm 2 2 2 2 2 As acceleration is the same as deceleration, the load torques are given by: M b load = M v load = J load load max = 0.00281 698.2 = 196 . Nm Selecting the motor The following motor is selected (refer to calculating the motor torques): 1FT6 061-6AC7, nn=2000 RPM, Jmotor=0.0006 kgm2 Mn(100)=3.7 Nm, M0(100)=4 Nm, I0(100)=2 A, motor=0.82 256 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks Calculating the motor torques when accelerating and decelerating Accelerating- and decelerating torque for the motor: M b motor = M v motor = J motor load max = 0.0006 698.2 = 0.42 Nm Motor torque when accelerating: M motor = M b motor + M b load = 0.42 + 196 . = 2.38 Nm Motor torque when decelerating: M motor = - M v motor - M v load = -0.42 - 196 . = -2.38 Nm (regenerative operation) The highest motor torque is required when accelerating and when decelerating. The selected motor is adequately dimensioned, as it has sufficient reserves at nmax=645.6 RPM and a 400 V supply voltage. The coupling moment of inertia was neglected. Further, no frictional forces were taken into account. Selecting the drive converter The following is valid for the highest required motor torque: M motor max M 0 (100) = 2.38 = 0.6 4 The following value is obtained from the torque/current characteristic: I motor max I 0 (100) = 0.6 and I motor max = I 0 (100) 0.6 = 2 0.6 = 12 . A Selected drive converter: 6SE7011-5EP50 (Compact Plus) PV n=0.55 kW; IV n=1.5 A, IV max=2.4 A (160% overload capability) Siemens AG SIMOVERT MASTERDRIVES - Application Manual 257 3 Various special drive tasks 09.99 Dimensioning the brake resistor The maximum braking power for the brake resistor when decelerating from load max to load min: Pbr W max = M motor v load max motor = 2.38 67.6 0.82 = 132 W Braking power for the brake resistor at load min: Pbr W ( load min ) = M motor v load min motor = 2.38 22.22 0.82 = 43.4 W Braking diagram P br W 132 W 43.4 W tv t ts Braking energy for a cycle (corresponds to the area in the braking diagram): Wbr = = 258 Pbr W max + Pbr W ( load min ) 2 tv 132 + 43.4 0.065 = 5.7 Ws 2 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks The following must be valid for the brake resistor: Wbr 0.0057 = = 0.038 kW Pbr cont . ts 0.15 with Pbr cont. = P20 4.5 (with an external brake resistor) the following is obtained 4.5 0.038 = 0171 . kW P20 Thus, the smallest braking resistor is selected with P 20 = 5 kW (6SE7018-0ES87-2DC0). Thermally checking the motor Torque characteristic M motor 2.38 Nm 0.065 s t 0.065 s -2.38 Nm 0.15 s The RMS torque is obtained from the torque characteristic: M RMS = 2 2.38 2 0.065 = 2.22 Nm 015 . Siemens AG SIMOVERT MASTERDRIVES - Application Manual 259 3 Various special drive tasks 09.99 With the following nload min = load min 60 22.22 60 = = 212.2 RPM 2 2 the average motor speed is given by: naverage = = nload min ( t s - 2 t b ) + nload min + nload max 2 2 tb ts 212.2 ( 015 . - 2 0.065) + ( 212.2 + 645.6) 0.065 = 400 RPM 015 . The calculated RMS torque is, at naverage, below the S1 curve. Thus, operation is thermally permissible. 260 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3.11 3 Various special drive tasks Centrifugal drive This involves a centrifugal drive with a specified load duty cycle. A 6-pole 250 kW induction motor is to be used (force-ventilated). Drive data Moment of inertia of the empty drum JL = 565 kgm2 Moment of inertia of the full drum JV = 1165 kgm2 Moment of inertia of the drum, water removed JE = 945 kgm2 Filling speed nF = 180 RPM Centrifuging speed nS = 1240 RPM Discharge speed nR = 70 RPM Filling time tF = 34 s Centrifuging time tS = 35 s Discharge time tR = 31.83 s Accelerating time from nF to nS tbS = 45 s Braking time from nS to nR tv = 39 s Discharge torque MR = 500 Nm Frictional torque Mfriction = 120 Nm Rated motor output Pn = 250 kW Rated motor current In = 430 A Rated motor torque Mn = 2410 Nm Rated motor speed nn = 989 RPM Motor efficiency motor = 0.96 Motor moment of inertia Jmotor Siemens AG SIMOVERT MASTERDRIVES - Application Manual = 7.3 kgm2 261 3 Various special drive tasks 09.99 Speed characteristic required for a cycle n nS 4 3 5 1 n F n R 6 2 t T Range 1: Accelerating from the discharge- to the filling speed t1=tbF Range 2: Filling at the filling speed t2=tF Range 3: Accelerating from the filling- to the centrifuging speed t3=tbS Range 4: Centrifuging speed t4=tS Range 5: Braking from the centrifuging- to the discharge speed t5=tv Range 6: Discharging at the discharge speed t6=tR Times tF, tbS, tS, tv and tR are specified. Acceleration from nR to nF should be realized with the same increase is speed as when accelerating to the centrifugal speed. Thus, the following is obtained for tbF: t bF = t bS nF - nR 180 - 70 = 45 = 4.67 s nS - n F 1240 - 180 The cycle time is given by: T = t bF + t F + t bS + t S + t v + t R = 4,67 + 34 + 45 + 35 + 39 + 3183 . = 189.5 s 262 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks It is assumed that the moment of inertia linearly increases and decreases during the filling phase (range 2), the accelerating phase to the centrifuging speed (range 3) and the discharge phase (range 6). Thus, the following moment of inertia characteristic is obtained over a cycle. J JV JE 1 JL 2 3 4 5 6 t T The following is generally true for the drive torque for changing speed and changing moment of inertia: M = J 2 dn n dJ + 60 dt 60 dt [Nm] J in kgm2, n in RPM Using the linear characteristics of curves n(t) and J(t), the following are obtained for the derivatives: dn n = dt t dJ J = dt t Siemens AG SIMOVERT MASTERDRIVES - Application Manual 263 3 Various special drive tasks 09.99 Calculating the motor torques The motor torques in each range are calculated, at the start (index 0) and the end (index 1). The characteristic is linear between these two points. Range 1, accelerating from the discharge- to the filling speed (n 0, J=0) M motor 10 = M motor 11 = ( J L + J motor ) = (565 + 7.3) 2 n F - n R + M friction 60 t bF 2 180 - 70 + 120 = 1532 Nm 60 4.67 Range 2, filling at the filling speed (n=0, J 0) M motor 20 = M motor 21 = = n F JV - J L + M friction 60 tF 180 1165 - 565 + 120 = 286 Nm 60 34 Range 3, accelerating, from the filling- to the centrifuging speed (n 0, J 0) M motor 30 = ( JV + J motor ) = (1165 + 7.3) 2 1240 - 180 180 945 - 1165 + + 120 = 2966 Nm 60 45 60 45 M motor 31 = ( J E + J motor ) = ( 945 + 7.3) 2 n S - n F n F J E - J V + + M friction 60 60 t bS t bS 2 n S - n F n S J E - JV + + M friction 60 60 t bS t bS 2 1240 - 180 1240 945 - 1165 + + 120 = 2152 Nm 60 45 60 45 Range 4, centrifuging at the centrifuging speed (n=0, J=0) M motor 40 = M motor 41 = M friction = 120 Nm Range 5, braking from the centrifuging- to the discharge speed (n 0, J=0) M motor 50 = M motor 51 = ( J E + J motor ) = ( 945 + 7.3) 264 2 n R - n S + M friction 60 tv 2 70 - 1240 + 120 = -2872 Nm 60 39 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks Range 6, discharging at the discharge speed (n=0, J 0) M motor 60 = M motor 61 = = nR J L - J E + M friction + M R 60 tR 70 565 - 945 + 120 + 500 = 576 Nm 60 3183 . Torque characteristic for a cycle M motor / Nm 3000 2000 1 3 1000 2 4 6 0 20 40 60 80 100 120 140 160 180 200 t/s -1000 5 -2000 -3000 Thermally checking the motor The motor enters the field-weakening range above the rated speed. In this range, a higher current for the same torque is required than in the constant-flux range. In order to thermally check the motor, the RMS motor torque is not used, but instead, the RMS motor current. Siemens AG SIMOVERT MASTERDRIVES - Application Manual 265 3 Various special drive tasks 09.99 The motor current is calculated in the constant-flux range as follows: I motor ( = const ) = ( M motor 2 2 2 2 ) ( I motor n - I n ) + I n M motor n for n nn The following is obtained in the field-weakening range: I motor ( field ) = ( n motor n 2 M motor 2 n motor 2 2 2 2 ) ( I motor - I ) ( ) + I ( ) n n n M motor n n motor n n motor for n nn Thus, the motor currents in each range, are calculated at the start (index 0) and at the end (index 1) and in ranges 3 and 5 in addition to the start of the field-weakening range (index FW). In ranges 3 and 5, a simplification has been made by assuming that the motor current has a linear characteristic between the two points. The field-weakening range is obtained from the following speed diagram. Field weakening range n t 3 FW t 5 FW nS nn 3 n 4 5 F n R t t bS t 3 FW = t bS t 5 FW = t v 266 tS tv n S - nn 1240 - 989 = 45 = 10.66 s 1240 - 180 nS - nF n S - nn 1240 - 989 = 39 = 8.37 s 1240 - 70 nS - n R Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks Range 1, accelerating from the discharge- to the filling speed M motor 10 = M motor 11 = 1532 Nm I motor 10 = I motor 11 = ( 1532 2 ) (430 2 - 133.3 2 ) + 133.32 = 292 A 2410 Range 2, filling at the filling speed M motor 20 = M motor 21 = 286 Nm I motor 20 = I motor 21 = ( 286 2 ) (430 2 - 133.3 2 ) + 133.3 2 = 142 A 2410 Range 3, accelerating from the filling- to centrifuging speed (field-weakening range from nn to nS) M motor 30 = 2966 Nm I motor 30 = ( 2966 2 ) (430 2 - 133.3 2 ) + 133.32 = 520 A 2410 M motor 3 FW = M motor 31 - I motor 3 FW = ( M motor 31 - M motor 30 t bS t 3 FW = 2152 - 2152 - 2966 10.66 = 2344.8 Nm 45 2344.8 2 ) (430 2 - 133.32 ) + 133.32 = 419.5 A 2410 M motor 31 = 2152 Nm I motor 31 = ( 2152 2 1240 2 989 2 ) (430 2 - 133.32 ) ( ) + 133.32 ( ) = 469.9 A 1240 2410 989 Range 4, centrifuging at the centrifuging speed (field-weakening range) M motor 40 = M motor 41 = M friction = 120 Nm I motor 40 = I motor 41 = ( 120 2 1240 2 989 2 ) (430 2 - 133.3 2 ) ( ) + 133.3 2 ( ) = 109.3 A 2410 989 1240 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 267 3 Various special drive tasks 09.99 Range 5, braking from the centrifuging- to the discharge speed (field-weakening range from nS to nn) M motor 50 = M motor 5 FW = M motor 51 = -2872 Nm I motor 50 = ( 2872 2 1240 2 989 2 ) (430 2 - 133.3 2 ) ( ) + 133.3 2 ( ) = 620 A 2410 989 1240 I motor 5 FW = I motor 51 = ( 2872 2 ) (430 2 - 133.3 2 ) + 133.32 = 505 A 2410 Range 6, discharge at the discharge speed M motor 60 = M motor 61 = 576 Nm I motor 60 = I motor 61 = ( 576 2 ) (430 2 - 133.32 ) + 133.32 = 165.3 A 2410 Current characteristic for a cycle I motor / A Field weakening range 700 600 500 400 300 3 5 1 200 100 2 4 6 0 0 268 20 40 60 80 100 120 140 160 180 200 t/s Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks The RMS motor current is calculated from the current characteristic. For non-constant currents in ranges 3 and 5, the following is obtained for an interval: i +1 I 2 dt = i 1 ( I i2 + I i2+1 + I i I i +1 ) ti 3 Thus, the following is true: 2 2 2 I motor RMS T = I10 t bF + I 20 t F + 1 2 ( I30 + I32 FW + I30 I3 FW ) (tbS - t3 FW ) 3 1 2 + ( I32FW + I312 + I3 FW I31 ) t3 FW + I40 tS 3 1 + ( I502 + I52 FW + I50 I5 FW ) t5 FW + I52 FW (tv - t5 FW ) + I602 t R 3 1 = 292 2 4.67 + 142 2 34 + (520 2 + 419.5 2 + 520 419.5) (45 - 10.66) 3 1 + (419.52 + 469.9 2 + 419.5 469.9) 10.66 + 109.32 35 3 1 + (620 2 + 5052 + 620 505) 8.37 + 5052 ( 39 - 8.37) + 165.32 3183 . 3 = 22557472 A 2 s and I motor RMS = 22557472 = 345 A 189.5 This RMS motor current is permissible as the rated motor current is 430 A. Calculating the braking power Braking occurs when decelerating from the centrifuging speed to the discharge speed. In this case, the drive is braked with a constant motor torque of 2872 Nm. Max. braking power in the DC link at nS: Pbr DClink max = M motor v n S 9550 motor WR = 2872 1240 0.96 0.98 = 350.8 kW 9550 Minimum braking power in the DC link at nR: Pbr DClink min = M motor v n R 9550 motor WR = Siemens AG SIMOVERT MASTERDRIVES - Application Manual 2872 70 0.96 0.98 = 19.8 kW 9550 269 3 Various special drive tasks 09.99 Braking power characteristic for a cycle P br DC link P br DC link max 5 P br DC link min t tv Braking energy for a cycle (corresponds to the area in the braking diagram): Wbr = Pbr DClink max + Pbr DClink min 2 tv = 350.8 + 19.8 39 = 7226.7 kWs 2 The following is obtained for the average braking power: Pbr DClink average = Wbr 7226.7 = = 38 kW T 189.5 As a result of the high average braking power, a brake resistor is not used, but instead, a drive converter with regenerative feedback into the line supply. Selecting the drive converter The maximum motor current is 620 A (braking mode) and the RMS motor current is 345 A. Thus, the following drive converter is selected with regenerative feedback into the line supply. 6SE7135-1EF62-4AB0 PV n=250 kW; IV n=510 A, IV max=694 A, IDC link n=605 A 270 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks It must still be checked whether the maximum DC link current when braking, in regenerative operation, is permissible. I DClink gen max = Pbr DClink max U ZK = Pbr DClink max 1.35 Vsup ply = 350.8 10 3 = 650 A 1.35 400 The permissible DC link current in regenerative operation is 92% of the value permissible for motor operation: I DClink gen perm. = I DClink n 136 . 0.92 = 605 136 . 0.92 = 757 A Thus, the infeed/regenerative feedback unit is adequately dimensioned. Selecting the regenerative feedback transformer When selecting the regenerative feedback transformer, the RMS DC link current in regenerative feedback operation is first calculated. I DClink gen = Pbr DClink VDClink The RMS value in regenerative feedback operation is obtained using the following equation: I DClink gen RMS = Pbr DClink RMS VDClink = 1 VDClink 1 ( Pbr2 DClink max + Pbr2 DClink min + Pbr DClink max Pbr DClink min ) t v 3 T 1 (350.8 2 + 19.8 2 + 350.8 19.8) 39 1 = 3 = 175 A 540 189.5 The permissible RMS value for a regenerative feedback transformer with 25% duty ratio factor is given by: I DClink RMS perm. = I DClink n 0.92 25 = I DClink n 0.46 = 605 0.46 = 278.3 A 100 Thus, a regenerative feedback transformer 4BU5195-0UA31-8A, with 25% duty ratio factor is adequate. For several centrifugal drives, a multi-motor drive can be configured, consisting of a rectifier/regenerative feedback unit, a DC link bus and an inverter for each centrifuge. Energy is transferred through the DC link bus by appropriately harmonizing the individual cycles. Thus, a lower-rating rectifier/regenerative feedback unit can be selected. Siemens AG SIMOVERT MASTERDRIVES - Application Manual 271 3 Various special drive tasks 3.12 09.99 Cross-cutter with variable cutting length For a force-ventilated induction motor, the possible web speed should be defined as a function of the cut length. The mechanical configuration is as follows. v web IM Gearbox knife rolls Drive data Diameter of the knife rolls D = 284 mm Moment of inertia of the knife rolls Jload = 6.43 kgm2 Mechanical efficiency mech = 0.95 Overlap angle u = 60 degrees Max. permissible web velocity vB per. = 349 m/min Max. permissible circumferential velocity of the knife rolls vU W per. = 363 m/min Min. cutting length sL min = 372 mm Max. cutting length sL max = 1600 mm Gearbox ratio i = 2.029 Supplementary moment of inertia, motor side Jmot suppl. = 0.18 kgm Cutting torque MS = 115 Nm Rated motor output Pmot n = 90 kW Rated motor torque Mmot n = 1150 Nm Motor stall torque Mstall = 3105 Nm Rated motor speed nmot n = 750 RPM Motor moment of inertia Jmot = 0.67 kgm2 272 2 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks Rated motor current Imot n = 200 A Magnetizing current Imag = 97 A Efficiency mot = 0.92 Rated motor voltage Umot n = 340 V When defining the possible web velocity as a function of the cut length, the following limits must be observed. * Permissible limiting torque for the motor (Mmot max Mper.) * Thermal limit for the motor (Irms Imot n) * Max. permissible web velocity * Max. permissible circumferential velocity of the knife rolls The maximum motor torque when accelerating is obtained from M Mot max = M Mot b = ( J Mot + J Mot sup pl . ) i b + J load b b 1 i mech angular acceleration of the knife rolls When decelerating, the motor torque is given by v = - b M Mot v = - ( J Mot + J Mot sup pl . ) i b - J load b mech i Precisely during the overlap phase (constant motor speed), the motor should provide the cutting torque. M Mot k = MS i mech The motor current is obtained from the motor torque: I Mot I Mot n ( kn = 1 kn = n Mot n Mot n I mag 2 I mag 2 1 M Mot 2 ) (1 - ( ) ) kn 2 + ( ) 2 M Mot n I Mot n I Mot n kn for n Mot n Mot n constant flux range for n Mot > n Mot n field weakening Siemens AG SIMOVERT MASTERDRIVES - Application Manual 273 3 Various special drive tasks 09.99 In order to calculate the characteristic of the motor torque and motor current over time as well as the RMS motor current, the motion sequences of the knife rolls must first be defined as a function of the cut length. Motion sequences of the knife rolls For the specified overlap angle u , the web speed and the circumferential velocity of the knife rolls should be in synchronism. B D u v B B u overlap angle vB constant web velocity B = vB D 274 2 D angular velocity of the knife rolls during the overlap diameter of the knife rolls Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks Depending on the cut length sL , for the remaining angular range, the following different situations are relevant for the angular velocity characteristic of the knife rolls. 1. Sub-synchronous lengths ( sL < D ) For sL < D , the cut length is less than the circumference of the knife rolls. This means, it must be appropriately accelerated to "catch-up" and then decelerated. The following diagram is obtained. d cut B t _k 2 tb _t k tv 2 ts d tk = angular velocity of the knife rolls at the reversal point u B tb = t v = time interval with constant angular velocity B d -B b b ts = accelerating time, decelerating time of the knife rolls angular acceleration of the knife rolls sL 2 sL = = t + 2 tb vB B D k time between 2 cuts For one revolution of the knife rolls, the area under the curve in the diagram is given by: 2 = B t s + ( d - B ) t b Siemens AG SIMOVERT MASTERDRIVES - Application Manual 275 3 Various special drive tasks 09.99 By inserting this, the following equations are obtained for the quantities which are required. tb = 2 - 2 sL D b sL u - B = D 2 tb d = B + 2 - 2 sL D tb If B is known, then times t k and t s can be calculated with the previously specified formulas. The angular acceleration b is then obtained as follows from the maximum motor torque when accelerating, b = M Mot max i ( J Mot + J Mot sup pl . ) + J load i mech The maximum motor speed is given by: n Mot max = d 60 i 2 The following condition must be maintained for sL min : sL min > u D 2 2. Synchronous length ( sL = D ) For sL = D (the cut length corresponds to the circumference of the knife rolls), d = B . This means, that there is neither acceleration nor deceleration. The web velocity could theoretically be set as required. However, in practice it is limited by v B per. . 276 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks 3. Above-synchronous lengths ( sL > D ) For sL > D , the cut length is greater than the circumference of the knife rolls. Therefore, after the cut, the system must be braked and then re-accelerated. This results in the following diagram. cut B d tk 2 tb tv ts tk t 2 In this case, tb = t v = B -d b Thus, the acceleration time is given by: tb = 2 sL - 2 D b The maximum motor speed is now defined by B : n Mot max = B 60 i 2 All other equations correspond to those of the sub-synchronous length. Siemens AG SIMOVERT MASTERDRIVES - Application Manual 277 3 Various special drive tasks 09.99 In the limiting case, d = 0 . The following is valid for the associated cut length: s L lim it = 2 D - u D 2 If the cut length is increased even further, an interval must be inserted with d = 0 . Thus, the following diagram is obtained cut B tk 2 tb tp tv tk t 2 ts For the area under the curve in the diagram, the following is obtained for one revolution of the knife rolls: 2 = u + B tb The following is still true b = B tb Thus, the following is obtained: tb = 2 - u b B = 2 - u tb t p = ts - tk - 2 tb interval for d = 0 All of the other equations correspond to those of the sub-synchronous lengths. 278 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks Making the calculation The calculation is made using an Excel/VBA program for a specific cut length according to the following schematic: 1) Enter Mmot max and calculate the motor RMS current 2) Interrogate Irms Imot n , when required, iterate up to Irms = Imot n 3) Interrogate Mmot max Mstall (nmot max), when required, iterate up to Mmot max = Mstall(nmot max) 4) Interrogate vB vB per. , when required limit 5) Interrogate vU W max vU W per. , when required limit To Point 1) Initially, Mmot max is set to Mper. (e.g. 2 Mmot n). The angular acceleration b is obtained with Mmot max. Depending on the range (sub-synchronous, above-synchronous, above-synchronous with sL>sL limit), the quantities vB, nmot max, tk, tb, ts and tp quantities are defined. If nmot max > nmot n , the instant where the field weakening becomes active must be calculated for the current calculation. Naturally, nmot max may not exceed the permissible limiting speed of the motor. If an interval tp occurs for above-synchronous length, then the motor torque is set to zero in this range. The characteristic of the motor torque with respect to time, and therefore also the motor RMS current, can be calculated. Siemens AG SIMOVERT MASTERDRIVES - Application Manual 279 3 Various special drive tasks 09.99 nMot nMot max Field weakening range nMot n t FW = nMot B 2 tFW n Mot max - n Mot n n Mot max - n Mot B tb t tk 2 tk tv tb 2 MMot MMot b MMot k t MMot v IMot IMot max t 0 1 2 3 4 5 6 Example for motor speed, motor torque and motor current for sub-synchronous lengths As a result of the efficiency when accelerating, the motor torque is greater than when decelerating, and the motor current when accelerating is correspondingly higher. The current increases in the field-weakening range. The RMS value is obtained for the motor torque and the motor current with the formulas specified previously: I rms = i =6 I Mot i -1 + I Mot i i =1 2 ( ) 2 (t i - t i -1 ) ts For above-synchronous lengths with s L > s L lim it , there is also an interval with Mmot=0. 280 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks To Point 2) If Irms Imot n , then continue with Point 3). For Irms > Imot n , Mmot max is reduced until Irms = Imot n. This is an iterative procedure. After this, the calculation continues with Point 3). To Point 3) First, a check is made whether nmot max nlimit (nlimit is the application point for the stall limit, 2 ). decreasing with 1 / n Mot nlim it = n Mot n M stall M per . 1,3 M per. M stall n Mot n 2 ) ( n Mot 1,3 n limit n Mot If nmot max nlimit, the calculation continues with Point 4). Otherwise, the stall limit is checked: M Mot max n Mot n 2 M stall ( ) n Mot max 13 . If this condition is fulfilled, continue with Point 4). Otherwise, Mmot max is reduced until the stall limit is just reached. This is an iterative procedure. Then continue with Point 4). Siemens AG SIMOVERT MASTERDRIVES - Application Manual 281 3 Various special drive tasks 09.99 To Point 4) If vB vB per., then continue with Point 5). Otherwise, vB is limited to vB per.. The remaining values (d, tk, tb, ts, tp) must be appropriately converted according to the reduced vB. With B per . = v B per . 2 D the remaining values are obtained: d = B per . (1 + 2 sL D ) sL u - D 2 2 - sL u - 2 D tb = B per . tk = u B per . ts = 2 sL B per . D or d = 0 or t b = tp = 0 2 - u B per . or t p = t s - t k - 2 t b for s L > s L lim it for s L > s L lim it for s L > s L lim it Due to the fact, that when vB is limited, the maximum circumferential velocity vU W max of the knife rolls can no longer increase, the interrogation operations are therefore completed. To Point 5) For above-synchronous lengths, the interrogation operations have been completed, as, in this case, the maximum circumferential velocity of the knife rolls is defined by vB. For sub-synchronous lengths, the maximum circumferential velocity of the knife rolls is defined by d: vU W max = d D 2 For vU W max vU W per., the interrogation operations have been completed. Otherwise, vU W max is limited to vU W per.. The remaining values (B, tk, tb, ts) must be converted corresponding to the reduced vU W max. With d per . = vU W per . 282 2 D Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks the remaining values are obtained: 1 B = d per . 1+ 2 sL D sL u - D 2 2 - sL u - tb = D 2 B tk = u B ts = 2 sL B D Siemens AG SIMOVERT MASTERDRIVES - Application Manual 283 3 Various special drive tasks 09.99 Dimensioning a capacitor battery A capacitor battery can be used for buffering, to prevent power oscillations between the motor and line supply. This capacitor battery is dimensioned, so that in operation, energy is neither fed back into the line supply nor converted into heat in the pulsed resistor. Without taking into account the losses, the regenerative feedback energy from the kinetic energy is obtained as follows: Wgen = Wkin = 1 J tot . wd2 - w B2 2 sub-synchronous and above-synchronous with J tot . = J load + ( J Mot + J Mot sup pl . ) i 2 Using the capacitor energy WC = 1 C (U d2 max - U d2 n ) 2 the following is obtained for the capacitor: C= 2 Wkin max U d2 max - U d2 n The maximum kinetic energy generally occurs at sL min. When dimensioning the capacitor battery, the existing capacitance in the drive converter DC link or inverter, must be taken into account. If one takes into account the efficiencies mech, mot, WR, the regenerative feedback energy must be calculated using the negative area under the curve of the DC link power. Wgen = PDClink gen dt = 1 M Mot v ( d + B ) Mot WR i t v 2 After several rearrangements, the following is obtained: Wgen = 1 J tot . 2d - 2B Mot WR 2 with 2 J tot . = J load mech + ( J Mot + J Mot sup pl . ) i 284 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks Results as characteristics The following characteristics are obtained with the specified numerical values. 350 300 web velocity in m/min 250 200 150 100 50 0 300 500 700 900 1100 1300 1500 1700 Cut length in mm Web velocity vB 45 40 omega in 1/s 35 30 25 wB wd 20 15 10 5 0 300 500 700 900 1100 1300 1500 1700 Cut length in mm Angular velocities B and d with limits vB per. and vU W per. (intersection at the synchronous cut length sL syn = D = 892,2 mm ) Siemens AG SIMOVERT MASTERDRIVES - Application Manual 285 3 Various special drive tasks 09.99 1 0.9 0.8 Irms / I mot n 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 300 500 700 900 1100 Cut length in mm 1300 1500 1700 Motor RMS current referred to the rated motor current 0.7 0.6 M mot max / M mot n 0.5 0.4 0.3 0.2 0.1 0 300 500 700 900 1100 Cut length in mm 1300 1500 1700 Max. motor torque referred to the rated motor torque 286 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks 1.4 I mot max / I mot n 1.2 1 0.8 0.6 0.4 0.2 0 300 500 700 900 1100 Cut length in mm 1300 1500 1700 Max. motor current referred to the rated motor current 10 8 Energy in kWs 6 4 2 0 300 W mot W gen 500 700 900 1100 1300 1500 1700 -2 -4 -6 -8 Cut length in mm Energy when motoring and generating Siemens AG SIMOVERT MASTERDRIVES - Application Manual 287 3 Various special drive tasks 09.99 The following characteristics (over time) are obtained for 2 various cut lengths (sub-synchronous and above-synchronous). 400 350 300 v in m/min 250 200 150 100 50 0 0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2 Time in s Circumferential velocity of the knife rolls for sL=716 mm (sub-synchronous) The intervention point for field-weakening is v=329.8 m/min 1500 Motor torque in Nm 1000 500 0 0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2 -500 -1000 -1500 Time in s Motor torque for sL=716 mm (sub-synchronous) 288 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks 250 Current in A 200 150 100 50 0 0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2 Time in s Motor current for sL=716 mm (sub-synchronous) Current increase in the field-weakening range 120 100 DC link power in kW 80 60 40 20 0 0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2 -20 -40 -60 -80 Time in s DC link power for sL=716 mm (sub-synchronous) The negative area under the curve corresponds to the regenerative energy Siemens AG SIMOVERT MASTERDRIVES - Application Manual 289 3 Various special drive tasks 09.99 350 300 v in m/min 250 200 150 100 50 0 0 0.02 0.04 0.06 0.08 0.1 Time in s 0.12 0.14 0.16 0.18 Circumferential velocity of the knife rolls for sL=1016 mm (above-synchronous) The intervention for field-weakening is v=329.8 m/min 800 600 Motor torque in Nm 400 200 0 0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 -200 -400 -600 -800 Time in s Motor torque for sL=1016 mm (above-synchronous) 290 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks 160 140 Current in A 120 100 80 60 40 20 0 0 0.02 0.04 0.06 0.08 0.1 Time in s 0.12 0.14 0.16 0.18 Motor current for sL=1016 mm (above-synchronous) Current increase in the field-weakening range 80 DC link power in kW 60 40 20 0 0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 -20 -40 -60 Time in s DC link power for sL=1016 mm (above synchronous) The negative area under the curve corresponds to the regenerative energy Siemens AG SIMOVERT MASTERDRIVES - Application Manual 291 3 Various special drive tasks 09.99 Drive converter selection The drive converter is selected according to the maximum motor current and the maximum RMS value of the motor current for all cut lengths. The maximum RMS value is exactly the same as the rated motor current (200 A). The maximum motor current is obtained when accelerating and at the maximum motor speed (in the field-weakening range). The maximum value is at precisely the minimum cut length with Imot max=259.2 A. Thus, the following SIMOVERT MASTERDRIVES MOTION CONTROL drive converter is obtained: 6SE7032-6EG50 PV n=110 kW IV n=218 A IV max=345 A Selecting the additional capacitor battery The maximum regenerative energy occurs in this case, at the minimum cut length. Wgen max = 6,77 kWs Thus, the required capacitance is given by C= 2 Wgen max U 2 d max -U 2 dn = 2 6770 = 73 mF 6752 - 520 2 The DC link capacitance of the drive converter is 14.1 mF. This means that an additional capacitance of 58.9 mF is required. The pre-charging of the external capacitor battery is realized after internal charging has been completed, using a separate pre-charging circuit. 292 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks 3.13 Saw drive with crank Stacks of paper napkins should be cut using a series of circular saw blades, connected to a crank pinion. The saw blades, driven using a toothed belt, are moved in a vertical direction via the crank drive. Saw blades Belt drive Gearbox M Motor Mode of operation Drive data Weight of the saw blades msaw = 80 kg Weight of the crank pinion mpinion = 13.5 kg Weight of the connecting rod mrod = 6.5 kg Siemens AG SIMOVERT MASTERDRIVES - Application Manual 293 3 Various special drive tasks 09.99 Weight of the crank disk mdisk = 10 kg Length of the crank pinion r2 = 1.25 m Length of the connecting rod l = 0.6 m Radius of the crank disk r = 0.12 m point of rotation r1 = 0.8 m Cycle time T = 0.625 s Clearance between the center of the saw blades and the fixed Motion sequence Starting from the upper dead center of the cam disk, it accelerates through 180 degrees to the lower dead center, followed by deceleration through 180 degrees back to the upper dead center followed by a no-load interval. Upper dead center b a Lower dead center The 180 degree angular range between points a and b should be traveled through within 160 ms. The cycle time is 625 ms. The characteristics for the angular velocity and the angular acceleration of the crank disk are as follows. 294 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks max 2 2 t 160 ms t total 625 ms max t - max max and ttot are given by: max = 2 2 = 23 s -1 0,16 1 + 2 t tot . = 016 . (2 + 2 ) = 0.546 s Thus, the angular acceleration of the crank disk is given by: max = max 23 2 = = 84.2 s -2 t tot . 0.546 2 The maximum crank disk speed is given by: nmax = max 60 23 60 = = 219.6 RPM 2 2 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 295 3 Various special drive tasks 09.99 The angular acceleration , the angular velocity and the angle of the crank disk can now be represented as a function of time. The zero point of the angle is at the lower dead center. Acceleration range: 0 t t tot . or - 0 2 = max = max t 1 = - + max t 2 2 Deceleration range: t tot . t t tot . or 0 2 = - max = max (t tot . - t ) 1 = - max (t tot . - t ) 2 2 The following simplifications were made when calculating the forces and torques. A m rod 2 m saw Crank pinion r1 0 r2 l Connecting rod B m rod 2 Crank disk The weight of the saw blades including the mounting, are assumed to act at a single point at a distance r1 from the fixed point of rotation 0. The weight of the connecting rod mrod is distributed evenly between points A and B. The crank pinion is assumed to be a thin bar with mass mpinion. The cutting torque of the saw blades is neglected. 296 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks Forces at the crank pinion With the previously specified simplifications, the force Fx in the x axis, acting at point A is obtained. This force consists of the accelerating component Fbx and a weight component FGx. The accelerating force Fbx acts in the opposite direction to acceleration ax. ax A mrod 2 msaw r1 0 r2 Fx The moment of inertia of this arrangement referred to the fixed point of rotation 0: J tot . = mrod 2 1 r2 + msaw r12 + m pinion r22 2 3 With acceleration ax, the accelerating force at point A is obtained as follows: Fbx r2 = J tot . ax r2 or Fbx = J tot . ax m r 1 = ( St + msaw ( 1 ) 2 + m pinion ) a x 2 r2 r2 2 3 The acceleration ax , initiated by the crank movement is, with and (also refer to Example 3.6), given by: r2 a x r ( cos + sin ) - ( 2 cos 2 + sin 2 ) l 2 2 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 297 3 Various special drive tasks 09.99 The following is valid for the force due to the weight at point A: FGx = ( mrod r 1 + msaw 1 + m pinion ) g 2 r2 2 Torques at the crank disk The torque at the crank disk due to force Fx (also refer to Example 3.6) is given by: M Fx = ( Fbx + FGx ) sin( - + ) r cos with r = arcsin( sin ) l In addition, a torque is effective due to the weight of half of the rod mass at Point B. r mrod 2 B M rod / 2 = 298 mrod g r sin 2 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks The moment of inertia of the crank disk is calculated as follows when assuming a full cylinder: J disk = 1 mdisk r 2 2 For half of the rod mass at point B, the moment of inertia is given by: J rod /2 = 1 mrod r 2 2 The total torque which must be provided at the crank disk: M crank = M Fx + M rod / 2 + ( J disk + J rod / 2 ) With the specified numerical values, the torque required at the crank disk can be calculated, e.g. using an Excel program. 150 Crank torque in Nm 100 50 0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 -50 -100 -150 Time in s Selecting the motor Selected motor: 1FT6044-4AF7 with gearbox i=10 (gearbox=0.95), nn=3000 RPM, Jmot=0.00062 kgm2 (with brake), mot=0.89, Jgearbox=0.000133 kgm2, kT0=1.67 Nm/A, M0=5 Nm, I0=3 A, Mmax=20 Nm, Imax=11 A Siemens AG SIMOVERT MASTERDRIVES - Application Manual 299 3 Various special drive tasks 09.99 Maximum motor speed: n Mot max = i n max = 10 219.6 = 2196RPM Motor torque The following is valid for the motor torque: M mot = J Mot i + J gearbox i + M crank 1 i VZ gearbox with VZ = sign( M crank ) 15 Motor torque in Nm 10 5 0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 -5 -10 -15 Time in s The highest motor torque is required during the upwards motion when motoring. This torque is 11.91 Nm. The motor torque during the downwards motion, in regenerative operation, is less as a result of the gearbox efficiency. 300 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks 20 18 M mot, M per. in Nm 16 14 12 10 8 6 4 2 0 0 500 1000 1500 2000 2500 3000 Motor speed in RPM The motor torque lies below the limiting characteristic. Thus, the motor is suitable as far as the dynamic limits are concerned. Thermally checking the motor The motor RMS torque is approximately given by: M rms = M 2 Mot dt T 2 2 M Mot i -1 + M Mot i + M Mot i -1 M Mot i 3 0.625 (t i - t i -1 ) = 4.5 Nm The average motor speed is obtained from: naverage = n Mot i -1 + n Mot i 2 T (t i - t i -1 ) n Mot max t tot . 2 2196 0.546 2 2 = = = 959.2 RPM 2 0.625 T Siemens AG SIMOVERT MASTERDRIVES - Application Manual 301 3 Various special drive tasks 09.99 5 4.5 Motor torque in Nm 4 3.5 3 M per. S1 M rms/n average 2.5 2 1.5 1 0.5 0 0 500 1000 1500 2000 Motor speed in RPM 2500 3000 The effective motor torque lies, at naverage, below the MS1 characteristic. This means that the motor is also suitable as far as the thermal limits are concerned. Dimensioning the brake resistor The motor output is obtained from the motor torque and the angular velocity as follows: PMot = M Mot i 2500 2000 1500 1000 Motor output in W 500 0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 -500 -1000 -1500 -2000 -2500 Time in s 302 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks The maximum braking power for the brake resistor is obtained from the maximum negative motor output (=motor braking power): Pbr W max = Pbr Mot max Mot WR = -2100 0,89 0,98 = -1832 W The braking energy for a cycle can be calculated using the negative area under the curve within the motor output characteristic. Wbr = PMot Mot WR dt Mot WR for PMot 0 PMot i-1 + PMot i 2 ( t i - t i-1 ) for PMot i -1 , PMot i 0 The zero positions of the motor output must be calculated using linear interpolation. A zero position between two points is obtained with the condition PMot i-1 PMot i < 0 to t x = t i-1 + t i - t i-1 P PMot i-1 - PMot i Mot i-1 The calculation results in: Wbr = -1238 . Ws The following is valid for the brake resistor: Wbr 1238 . = = 1981 . W Pbr duration T 0.625 With Pbr dur. = P20 4.5 (ext. brake resistor) the following is obtained 4.5 1981 . = 8914 . W P20 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 303 3 Various special drive tasks 09.99 The following must also be valid: Pbr W max = 1832 W 15 . P20 Thus, when selecting a compact type of drive converter, a brake unit with P 20=5 kW (6SE70180ES87-2DA0) and external brake resistor 5 kW (6SE7018-0ES87-2DC0) are required. When selecting a Compact Plus drive converter, only the external brake resistor is required (the chopper is integrated in the drive converter). Selecting the drive converter The drive converter is selected according to the peak value and the motor RMS current. The motor current is given by, taking into account saturation: I Mot = I Mot = M Mot kT0 b1 for M Mot M 0 M Mot M Mot - M 0 2 M I kT0 b1 (1 - ( ) (1 - max 0 )) M max - M 0 M 0 I max for M Mot > M 0 with b1 = 1 - b ( n Mot 1,5 ) 6000 (b=0.1 for frame size <100, otherwise 0.15) 7.13 A is obtained for the peak current. The RMS value is obtained from: I rms 2 2 I Mot i -1 + I Mot i + I Mot i -1 I Mot i 3 T (t i - t i -1 ) = 2.73 A Selected SIMOVERT MASTERDRIVES MOTION CONTROL drive converter: 6SE7015-0EP50 (Compact Plus type) PU n=1.5 kW; IU n=5 A; IU max=8 A with external 5 kW brake resistor (6SE7018-0ES87-2DC0) 304 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks 8.00 7.00 Motor current in A 6.00 5.00 4.00 3.00 2.00 1.00 0.00 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 Time in s Siemens AG SIMOVERT MASTERDRIVES - Application Manual 305 3 Various special drive tasks 3.14 09.99 Saw drive as four-jointed system The saw drive, investigated in 3.13, will now be re-calculated with modified data. The geometrical arrangement is such that the approximation with the crank disk, can no longer be used. Thus, this arrangement is known as the so-called four-jointed system. l pinion xr1 yr1 b pinion 3 yr2 2 xr2 Saw blades l yA 4 r 1 Crank disk xA Gearbox M Motor Mode of operation 306 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks Drive data Weight of the saw blades msaw = 21.5 kg Weight of the crank pinion mpinion = 26.5 kg Weight of the connecting rod mrod = 7.5 kg Moment of inertia of the crank disk Jdisk = 0.21 kgm2 Length of the crank pinion lpinion = 0.59 m Width of the crank pinion bpinion = 0.18 m Length of the connecting rod l = 0.4 m Radius of the crank disk r = 0.1 m xr 1 = 0.54 m fixed center of rotation 2 (y axis) yr 1 = 0.0325 m Clearance, point 3 to the fixed center of rotation 2 (x axis) xr 2 = 0.43 m Clearance, point 3 to the fixed center of rotation 2 (y axis) yr 2 = 0.035 m Clearance, point 1 to point 2 (x axis) xA = 0.35 m Clearance, point 1 to point 2 (y axis) yA = 0.5285 m Cycle time T =1s Clearance, center of the saw blades to the fixed center of rotation 2 (x axis) Clearance, center of the saw blades to the Motion sequence Starting from the upper dead center (i. e. connecting rod and crank pinion in one line), the crank disk rotates clockwise through 360 degrees within 0.5 s. This is then followed by an interval of 0.5 s. The maximum speed of the crank disk should be 192 RPM. Thus, the angular velocity and angular acceleration characteristics for the crank disk are as follows. Siemens AG SIMOVERT MASTERDRIVES - Application Manual 307 3 Various special drive tasks 1 09.99 2 1 max t tb tk tv t total T 1 1 max t - 1 max For 1 max, tb, tv and tk, the following is obtained: 2 n1 max 1 max = 60 tb = t v = = 2 192 = 20106 . s -1 60 1 max t tot - 2 20106 . 0.5 - 2 = = 01875 s . 20106 1 max . t k = t tot - 2 t b = 0.5 - 2 01875 . = 0125 . s Thus, the angular acceleration of the crank disk is given by: 1 max = 308 1 max 20106 . = = 107.232 s - 2 tb 01875 . Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks xr2 yr2 2 r 3 2 l yA 4 1 0 1 r xA Diagram to determine the starting angle at the upper dead center The starting angle at the upper dead center at t=0 is given by: 1 0 = arctan y A - yr 2 xr 2 - x A = arctan 0.5285 - 0.035 = 14101 . rad (corresponds to 80.79 degrees) 0.43 - 0.35 The angular acceleration 1, the angular velocity 1 and the angle 1 of the cam disk can now be represented as a function of time. Range: 0 t t b (acceleration) 1 = 1 max 1 = 1 max t 1 1 = 1 0 + 1 max t 2 2 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 309 3 Various special drive tasks 09.99 Range: t b t t b + t k (constant angular velocity) 1 = 0 1 = 1 max 1 1 = 1 0 + 1 max t b + 1 max ( t - t b ) 2 Range: t b + t k t t tot (deceleration) 1 = - 1 max 1 = 1 max - 1 max ( t - t b - t k ) 1 1 1 = 1 0 + 1 max t b + 1 max t k + 1 max ( t - t b - t k ) - 1 max ( t - t b - t k ) 2 2 2 Angular relationships in a four-jointed system r 3 2 2 2 3 l yA 4 r 1 1 xA 310 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks The following relationships are valid for angles 1, 2 and 3: l 2 = ( x A + r cos 1 - r2 cos 2 ) 2 + ( y A - r sin 1 + r2 sin 2 ) 2 tan 3 = Equation 1 x A + r cos 1 - r2 cos 2 y A - r sin 1 + r2 sin 2 Equation 2 with r2 = x r2 2 + y r2 2 Angle 1 is valid over the motion sequence. Angle 2 can be determined using Equation 1 (however, only iteratively). Angle 3 is obtained with 1 and 2 using Equation 2. Torques at the crank pinion In order to calculate the accelerating torques at the crank pinion, the moments of inertia with reference to the fixed center of rotation 2 must be known. The following simplifications are made. M2 b 2 Weight of the saw blades r1 b pinion r 3 l 2 2 pinion m rod 2 The weight of the saw blades including their mounting, is assumed to act at a single point at a distance r1 from the fixed center of rotation 2. The weight of the connecting rod mrod is distributed Siemens AG SIMOVERT MASTERDRIVES - Application Manual 311 3 Various special drive tasks 09.99 evenly between points 3 and 4. The crank pinion is assumed to be a square with length lpinion, width bpinion and with mass mpinion. The moment of inertia with respect to the fixed center of rotation 2 is then given by: 2 2 l pinion b pinion mrod 2 r + m pinion ( + ) J 2 = msaw r + 2 2 3 12 2 1 with r1 = x r21 + y r21 The accelerating torque with the angular acceleration of the crank pinion 2 is given by: M2 b = J 2 2 The angular velocity 2 of the crank pinion is obtained by implicitly differentiating Equation 1. 2 = r 1 ( A sin 1 + B cos 1 ) r2 ( A sin 2 + B cos 2 ) with A = x A + r cos 1 - r2 cos 2 B = y A - r sin 1 + r2 sin 2 The required angular acceleration 2 is obtained by differentiating 2: 2 = r ( 1 Z + 1 Z& ) - r2 2 N& r2 N with Z = A sin 1 + B cos 1 N = A sin 2 + B cos 2 Z& = sin 1 ( A& - B 1 ) + cos 1 ( B& + A 1 ) N& = sin 2 ( A& - B 2 ) + cos 2 ( B& + A 2 ) A& = - r 1 sin 1 + r2 2 sin 2 B& = - r 1 cos 1 + r2 2 cos 2 312 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks In order to calculate the torques due to the weight applied at the crank pinion, the angular difference to r2 must be known. The cutting torque of the saw blades is neglected. Weight of the saw blades r S M2 G 1 mpinion 3 2 mrod 2 r 1 = arctan S = arctan yr 2 xr 2 + arctan r2 r 1 yr 1 2 angle between r1 and r2 xr 1 yr 2 angle between the center of gravity axis and r2 xr 2 Thus, the torque due to the weight is given by: M 2 G = msaw g r1 cos( 2 + r 1 ) + mpinion g l pinion 2 cos( 2 + S ) + mrod g r2 cos 2 2 Calculating the connecting rod force The connecting rod force can be calculated using torques M2b and M2G. Siemens AG SIMOVERT MASTERDRIVES - Application Manual 313 3 Various special drive tasks 09.99 F pinion T r 3 2 M2 2 2 F rod 3 M2 = M2 b + M2 G FpinionT = Frod = torque around point 2 M 2b + M 2G tangential force vertical to r2 r2 Fpinion T cos( 3 - 2 ) = M2 b + M2 G r2 cos( 3 - 2 ) connecting rod force in direction l Torques at the crank disk The connecting rod force can be broken down into a tangential component vertical to r. 314 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks 3 Frod 4 mrod 2 F r crank T 1 mrod g 2 1 Fcrank T = Frod cos( 1 - 3 ) Thus, the torque as result of the connecting rod force is given by: M Frod = Fcrank T r = Frod r cos( 1 - 3 ) In addition, a torque is effective due to the weight of half of the connecting rod mass in Point 4. M mrod / 2 = mrod g r cos 1 2 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 315 3 Various special drive tasks 09.99 The moment of inertia for half of the connecting rod mass in Point 4 is given by: J mrod /2 = 1 m r2 2 rod Thus, the torque required at the crank disk is given by: M crank = M Frod + M mrod / 2 + ( J disk + J mrod / 2 ) 1 With the specified numerical values, the torque required at the crank disk can be calculated, e.g. using an Excel program. 200 150 Crank torque in Nm 100 50 0 0 0.2 0.4 0.6 0.8 1 -50 -100 -150 Time in s Selecting the motor Selected motor: 1FK6063-6AF71 with gearbox i=10 (gearbox=0.95), nn=3000RPM, Jmot=0.00167 kgm2 (with brake), mot=0.89, Jgearbox=0.000542 kgm2, kT0=1.33 Nm/A, M0=11 Nm, I0=8.3 A, Mmax=37 Nm, Imax=28 A Maximum motor speed: nmot max = i n1 max = 10 192 = 1920 RPM 316 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks Motor torque The following is valid for the motor torque: M mot = J mot 1 i + J gearbox 1 i + M crank 1 i VZ gearbox with VZ = sign( M crank ) 20 Motor torque in Nm 15 10 5 0 0 0.2 0.4 0.6 0.8 1 -5 -10 -15 Time in s The highest motor torque is required when moving upwards when motoring. This is 18.19 Nm. The motor torque when moving downwards when generating is lower due to the gearbox efficiency. During the pause interval at 1= 1 0, a holding torque is obtained due to 1 0 90 degrees. Only the weight of the connecting rod can participate in the holding torque, as the connecting rod and crank disk are in one line. This is: M mot hold = mrod 1 7.5 1 g r cos 1 0 = 9.81 01 . cos14101 . = 0.0589 Nm 2 i 2 10 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 317 3 Various special drive tasks 09.99 40 35 M mot, M per. in Nm 30 25 20 15 10 5 0 0 500 1000 1500 2000 2500 3000 Motor speed in RPM The motor torque lies below the limiting characteristic. Thus, the motor is suitable as far as the dynamic limits are concerned Thermally checking the motor The motor RMS torque is approximately given by: M RMS = M 2 mot dt T 2 2 M mot i -1 + M mot i + M mot i -1 M mot i 3 ( t i - t i -1 ) 1 = 5.71 Nm The average motor speed is obtained from: naverage = 318 nmot i-1 + nmot i 2 T ( t i - t i -1 ) nmot max = 2 t b 2 + nmot max t k T = + 0125 1920 ( 01875 . . ) = 600 RPM 1 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks 12 Motor torque in Nm 10 M per. S1 M RMS/n average 8 6 4 2 0 0 500 1000 1500 2000 2500 3000 Motor speed in RPM The effective motor torque lies, at naverage, below the MS1 characteristic. This means that the motor is also suitable as far as the thermal limits are concerned. Dimensioning the brake resistor The motor output is obtained from the motor torque and the angular velocity as follows: Pmot = M mot i 1 4000 3000 Motor power in W 2000 1000 0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 -1000 -2000 -3000 Time in s Siemens AG SIMOVERT MASTERDRIVES - Application Manual 319 3 Various special drive tasks 09.99 The maximum braking power for the brake resistor is obtained from the maximum negative motor output (=motor braking power): Pbr W max = Pbr mot max mot WR = -2379 0.89 0.98 = -2075 W The braking energy for a cycle can be calculated using the negative area under the curve within the motor output characteristic. Wbr = Pmot mot WR dt mot WR fur Pmot 0 Pmot i -1 + Pmot i 2 ( t i - t i-1 ) fur Pmot i -1 , Pmot i 0 The zero positions of the motor output must be calculated using linear interpolation. A zero position between two points is obtained with the condition Pmot i -1 Pmot i < 0 to t x = t i-1 + t i - t i -1 P Pmot i -1 - Pmot i mot i -1 The calculation results in: Wbr = -204.7 Ws The following is valid for the brake resistor: Wbr 204.7 = = 204.7 W Pbr duration T 1 With Pbr dur. = P20 4.5 (ext. Brake resistor) the following is obtained 4.5 204.7 = 921 W P20 320 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks The following must also be valid: Pbr W max = 2075 W 15 . P20 Thus, when selecting a compact type of drive converter, a brake unit with P 20=5 kW (6SE70180ES87-2DA0) and external brake resistor 5 kW (6SE7018-0ES87-2DC0) are required. When selecting a Compact Plus drive converter, only the external brake resistor is required (the chopper is integrated in the drive converter). Selecting the drive converter The drive converter is selected according to the peak value and the motor RMS current. The motor current is given by, taking into account saturation: I mot = I mot = M mot kT0 b1 for M mot M 0 M mot M mot - M 0 2 M I kT0 b1 (1 - ( ) (1 - max 0 )) M max - M 0 M 0 I max for M mot > M 0 with b1 = 1 - b ( nmot 1,5 ) 6000 (b=0.1 for frame size <100, otherwise 0.15) 13.93 A is obtained for the peak current. The RMS value is obtained from: I eff 2 2 I mot i -1 + I mot i + I mot i -1 I mot i 3 ( t i - t i-1 ) T = 4.35 A Selected SIMOVERT MASTERDRIVES MOTION CONTROL drive converter: 6SE7021-0EP50 (Compact Plus type) PU n=4 kW; IU n=10 A; IU max=16 A (160% overload capability) with external 5 kW brake resistor (6SE7018-0ES87-2DC0) Siemens AG SIMOVERT MASTERDRIVES - Application Manual 321 3 Various special drive tasks 09.99 14.00 12.00 Motor current in A 10.00 8.00 6.00 4.00 2.00 0.00 0 0.2 0.4 0.6 0.8 1 Time in s 322 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3.15 3 Various special drive tasks Mesh welding machine The drives for a mesh welding machine are to be dimensioned. In this case it involves the 4 drives: Trolley, roller feed, transverse wire feed and mesh transport. Synchronous servo motors with inverters, rectifier unit and central brake resistor are to be used. Trolley The trolley drive is a traversing drive to pull-in the longitudinal wires. A breakaway force is assumed at the start of pull-in for 200 ms, and is then further calculated with a steady-state pull-in force. When dimensioning the drive, there should be 100% safety margin regarding the breakaway force and pull-in force. Drive data Trolley mass mtrolley = 350 kg Load mass (wires) mload = 100 kg Breakaway force (for 200 ms) FLB = 2750 N Pull-in force, steady-state FZ = 600 N Frictional force FR = 400 N Travel s =6m Max. traversing velocity vmax = 2.5 m/s Traversing time, forwards ttot, forwards = 4 s Traversing time, backwards ttot, backwards = 3.25 s Pinion diameter D = 0.15152 m Clock cycle time T = 16 s Siemens AG SIMOVERT MASTERDRIVES - Application Manual 323 3 Various special drive tasks 09.99 Traversing characteristic v forwards backwards v max t t b for. t k for. t t tot back. b back. t k back. t v back. t v for. t tot for. - v max T a a max back. a max for. t - a max for. - a max back. t b for . = t v for . = t tot for . - s v max =4- 6 = 1 .6 s 2 .5 t k for . = ttot for . - 2 tb for . = 4 - 2 1.6 = 0.8 s a max for. = v max 2.5 = = 1.5625 m / s 2 tb for. 1.6 tb back . = t v back . = ttot back . - 324 s v max = 3.25 - 6 = 0.85 s 2 .5 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks t k back . = ttot back . - 2 tb back . = 3.25 - 2 0.85 = 1.55 s a max back . = v max tb back . = 2.5 = 2.94 m / s 2 0.85 Load-force characteristic F load 2F LB forwards backwards FR + 2 Ftension v max v LB t 0.2 s - FR T The safety margin of 100% for the breakaway force and the steady-state pull-in force are taken into account in the load-force characteristic. The traversing velocity until the breakaway force becomes effective, is given by: v LB = v max 0.2s 0.2 = 2.5 = 0.3125 m / s tb for. 1.6 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 325 3 Various special drive tasks 09.99 The following is valid for the load torque: M load for . = J load for . load + Fload D 2 M load back . = J load back . load + Fload D 2 with load = a J load for . 2 D D = (mtrolley + mload ) ( ) 2 2 J load back . = mtrolley ( D 2 ) 2 a from the characteristic a(t) Fload from the characteristic Fload(t) 500 400 Load torque in Nm 300 200 100 0 0 2 4 6 8 10 12 14 16 -100 -200 Time in s 326 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks Selecting the motor Selected motor: 2 1FT6082-8AK7 with gearbox i=16 (gearbox=0,95), nn=6000 RPM, Jmot=0,003 kgm , mot=0,88, Jgearbox=0,00082 kgm2, kT0=0,71 Nm/A, M0=13 Nm, I0=18,3 A, Mmax=42 Nm, Imax=73 A Maximum motor speed: nmot max = v max 60 i 2.5 60 16 = = 5042 RPM D 015152 . Motor torque The following is valid for the motor torque: M mot for . = J mot load i + J gear load i + M load for . 1 i VZ gear M mot back . = J mot load i + J gear load i + M load back . VZ gear i with VZ = sign( M load for .,back . ) 35 30 Motor torque in Nm 25 20 15 10 5 0 0 2 4 6 8 10 12 14 16 -5 -10 Time in s Siemens AG SIMOVERT MASTERDRIVES - Application Manual 327 3 Various special drive tasks 09.99 45 40 M mot, M per. in Nm 35 30 25 20 15 10 5 0 0 1000 2000 3000 4000 5000 6000 Motor speed in RPM The motor torque lies below the limiting characteristic. Thus, the motor is suitable as far as the dynamic limits are concerned. Thermally checking the motor The RMS motor torque is calculated from the motor torque characteristics m M RMS = 2 Mot i ti T = 616 . Nm The average motor speed is obtained from: naverage = = 328 nmot i -1 + nmoti 2 T ti nmot max = 2 tb for 2 + nmot max tk for + nmot max 2 tb back 2 + nmot max tk back T 5042 (16 . + 0.8 + 0.85 + 155 . ) = 1512.6 RPM 16 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks 14 Motor torque in Nm 12 10 M per. S1 M RMS / n average 8 6 4 2 0 0 1000 2000 3000 4000 5000 6000 Motor speed in RPM The effective motor torque lies, at naverage, below the MS1 characteristic. This means that the motor is also suitable as far as the thermal limits are concerned. Selecting the inverter The inverter is selected according to the peak value and RMS value of the motor current. Refer to example 3.13 for the current calculation. 49.6 A is obtained for the peak current. The RMS current is 9.23 A. In this particular case, the 300 % overload capability of the Compact Plus type of construction can be used, as the overload time is less than 250 ms; after the overload, the current returns to below 0.91 x of the rated current, and the recovery time is greater than 750 ms. Selected SIMOVERT MASTERDRIVES MOTION CONTROL inverter: 6SE7022-0TP50 (Compact Plus type of construction) Pinv n=11 kW; Iinv n=25.4 A; IDC link n=30.4 A; Iinv max=76.5 A (300 % overload capability) A 15 kW inverter would be considered if the 300 % overload capability was not to be used Iinv max=54.4 A (160 % overload capability). Siemens AG SIMOVERT MASTERDRIVES - Application Manual 329 3 Various special drive tasks 09.99 50.00 45.00 Motor current in A 40.00 35.00 30.00 25.00 20.00 15.00 10.00 5.00 0.00 0 2 4 6 8 10 12 14 16 Time in s Calculating the braking power and braking energy The motor output is obtained from the motor torque and the traversing velocity as follows: Pmot = M mot i 2v D 7000 6000 5000 Motor power in W 4000 3000 2000 1000 0 0 2 4 6 8 10 12 14 16 -1000 -2000 -3000 Time in s 330 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks The max. braking power for the brake resistor is obtained from the max. negative motor output (=motor braking power): Pbr W max = Pbr mot max mot inv = -2747.7 0.88 0.98 = -2370 W The braking energy for one cycle is obtained from the negative surface area under the motor output characteristic Wbr = 1 1 Pbr mot max t v down mot inv = - 2747.7 0.85 0.88 0.98 = -1007 Ws 2 2 The average braking power for one cycle is Pbr average = Wbr 1007 =- = -62.94 W 16 T Calculating the DC link currents when motoring The maximum DC link current and the RMS DC link current when motoring are required to dimension the rectifier unit. These currents are obtained from the positive motor output. The following is valid for the max. DC link current: I DC link max = Pmot max mot inv U DC link = 6726.4 = 14.44 A 0.88 0.98 1.35 400 The RMS DC link current is obtained as follows: I DC link RMS = 1 mot inv U DC link (P 2 mot i -1 2 + Pmot i + Pmot i Pmot i -1 ) T ti 3 for Pmot i -1 , Pmot i 0 A value of 3.76 A is obtained Siemens AG SIMOVERT MASTERDRIVES - Application Manual 331 3 Various special drive tasks 09.99 Roller feed For a roller feed drive, it involves a rotating drive to position the mesh to weld the transverse wires. A 50% safety margin for the feed torque should be taken into account. Drive data 2 Load moment of inertia Jload = 1.48 kgm Intrinsic moment of inertia JE = 0.35 kgm2 Feed torque Mv = 450 Nm Traversing angle per transverse wire = 5 rad Max. angular velocity max = 20 s-1 Traversing time per transverse wire ttot. = 0.45 s Cycle time per transverse wire T = 0.75 s Traversing characteristic max t tb tk tv t tot. T load load max - load max 332 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks tb = t v = ttot . - 5 = 0.45 - = 0 .2 s 20 max t k = ttot . - 2 tb = 0.45 - 2 0.2 = 0.05 s load max = max 20 = = 100 s -1 tb 0 .2 Load torque The load torque is given by: M load = ( J load + J E ) load + 1.5 M v load from the characteristic load(t) 900 800 Load torque in Nm 700 600 500 400 300 200 100 0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 Time in s Siemens AG SIMOVERT MASTERDRIVES - Application Manual 333 3 Various special drive tasks 09.99 Selecting the motor selected motor: 1FT6134-6SB. with gearbox i=7 (gearbox=0.95), nn=1500 RPM, Jmot=0.0547 kgm2, mot=0.94, Jgearbox=0.00471 kgm2, kT0=2.92 Nm/A, M0=140 Nm, I0=48 A, Mmax=316 Nm, Imax=139 A Maximum motor speed: nmot max = max 60 i 20 60 7 = = 1337 RPM 2 2 Motor torque The following is valid for the motor torque: M mot = J mot load i + J gearbox load i + M load 1 i VZ gearbox with VZ = sign( M load ) 180 160 Motor torque in Nm 140 120 100 80 60 40 20 0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 Time in s 334 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks 350 M mot, M per. in Nm 300 250 200 150 100 50 0 0 200 400 600 800 1000 1200 1400 1600 Motor speed in RPM The motor torque lies below the limiting characteristic. Thus, the motor is suitable as far as the dynamic limits are concerned. Thermally checking the motor The RMS motor torque is obtained from the motor torque characteristic as follows: M RMS = M 2 mot i ti T = 93.43 Nm The average motor speed is obtained as follows: naverage = nmot i -1 + nmot i 2 T ti nmot max = 2 Siemens AG SIMOVERT MASTERDRIVES - Application Manual t b 2 + nmot max t k T = 1337 ( 0.2 + 0.05) = 445.7 RPM 0.75 335 3 Various special drive tasks 09.99 140 Motor torque in Nm 120 100 80 M per. S1 M RMS / n average 60 40 20 0 0 200 400 600 800 1000 1200 1400 1600 Motor speed in RPM The effective motor torque lies, at naverage, below the MS1 characteristic. This means that the motor is also suitable as far as the thermal limits are concerned. Selecting the inverter 59.76 A is obtained for the peak current. The RMS value is 32.47 A. Selected SIMOVERT MASTERDRIVES MOTION CONTROL inverter: 6SE7023-8TP50 (Compact Plus type of construction) Pinv n=18.5 kW; Iinv n=37.5 A; IDC link n=44.6 A; Iinv max=60 A (160 % overload capability) 336 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks 60.00 Motor current in A 50.00 40.00 30.00 20.00 10.00 0.00 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 Time in s Calculating the braking power and the braking energy As can be seen from the motor torque characteristic, no braking power is obtained in normal operation. However, as braking without load and feed torque must also be taken into account, the braking power and braking energy are also calculated for this particular situation. The motor torque during the deceleration phase is given by: M mot v = -( J mot + J gearbox ) load max i - J E load max = -( 0.0547 + 0.00471) 100 7 - 0.35 100 gearbox i 0.95 = -46.34 Nm 7 The max. braking power for the brake resistor is given by: Pbr W max = M mot v max i mot WR = -46.34 20 7 0.94 0.98 = -5.98 kW Braking energy and average braking power for one cycle: Wbr = 1 1 Pbr W max t v = - 5.98 0.2 = -0.6 kWs 2 2 Pbr average = Wbr 600 =- = -800 W T 0.75 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 337 3 Various special drive tasks 09.99 Calculating the DC link currents when motoring The motor power is calculated from the motor torque and angular velocity as follows: Pmot = M mot i 25000 Motor power in W 20000 15000 10000 5000 0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 Time in s The following is valid for the max. DC link current: I DC link max = Pmot max mot inv U DC link = 23885 = 48.02 A 0.94 0.98 1.35 400 The DC link RMS current is given by: I DC link RMS 1 = mot inv U DC link (P 2 mot i -1 2 + Pmot i + Pmot i Pmot i -1 ) T ti 3 for Pmot i -1 , Pmot i 0 A value of 16.33 A is obtained 338 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks Transverse wire feed The transverse wire feed drive comprises two disks, which clamp the transverse wire. These disks are each moved from a motor in opposing directions to transport the transverse wire. Drive data Load moment of inertia per motor Jload = 0.0053 kgm2 Intrinsic moment of inertia per motor JE = 0.0463 kgm2 Feed torque per motor Mv = 180 Nm Traversing angle per transverse wire = 10.05 rad Max. angular velocity max = 29 s-1 Traversing time per transverse wire ttot. = 0.5 s Clock cycle time per transverse wire T = 0.75 s Traversing characteristic max t tb tk tv t tot. T load load max - load max Siemens AG SIMOVERT MASTERDRIVES - Application Manual 339 3 Various special drive tasks t b = t v = t tot - 09.99 10.05 = 0.5 - = 01534 . s max 29 t k = t tot - 2 t b = 0.5 - 2 01534 = 01931 s . . load max = max 29 = = 189 s -1 01534 tb . Load torque The load torque is given by: M load = ( J load + J E ) load + M v load from the characteristic load(t) 200 180 160 Load torque in Nm 140 120 100 80 60 40 20 0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 Time in s 340 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks Selecting the motor Selected motor: 2 1FT6086-8AF7. with gearbox i=7 (gearbox=0.95), nn=3000RPM, Jmot=0.00665 kgm , mot=0.91, Jgearbox=0.00174 kgm2, kT0=1.56 Nm/A, M0=18.5 Nm, I0=17.3 A, Mmax=90 Nm, Imax=72 A Maximum motor speed: nmot max = max 60 i 29 60 7 = = 1938.5 RPM 2 2 Motor torque The motor torque is given by: M mot = J mot load i + J gearbox load i + M load 1 i VZ gearbox with VZ = sign( M load ) 40 35 Motor torque in Nm 30 25 20 15 10 5 0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 Time in s Siemens AG SIMOVERT MASTERDRIVES - Application Manual 341 3 Various special drive tasks 09.99 90 80 M mot, M per. in Nm 70 60 50 40 30 20 10 0 0 500 1000 1500 2000 2500 3000 Motor speed in RPM The motor torque lies below the limiting characteristic. Thus, the motor is suitable as far as the dynamic limits are concerned. Thermally checking the motor The motor RMS torque is calculated from the motor torque characteristic as follows: M RMS = M 2 mot i ti T = 2352 . Nm The average motor speed is given by: naverage = 342 nmot i-1 + nmot i 2 T ti nmot max = 2 t b 2 + nmot max t k T = . . ) + 01931 1938.5 ( 01534 = 895.6 RPM 0.75 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks 30 Motor torque in Nm 25 20 15 M per. S1 M RMS / n average 10 5 0 0 500 1000 1500 2000 2500 3000 Motor speed in RPM The effective motor torque lies, at naverage, below the MS1 characteristic. This means that the motor is also suitable as far as the thermal limits are concerned. Selecting the inverter 26.1 A is obtained for the peak current. The RMS value is 15.34 A. Selected SIMOVERT MASTERDRIVES MOTION CONTROL inverter: 6SE7021-8TP50 (Compact Plus type) Pinv n=7.5 kW; Iinv n=17.5 A; IDC link n=20.8 A; Iinv max=28 A (160 % overload capability) Siemens AG SIMOVERT MASTERDRIVES - Application Manual 343 3 Various special drive tasks 09.99 30.00 Motor current in A 25.00 20.00 15.00 10.00 5.00 0.00 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 Time in s Calculating the braking power and the braking energy As can be seen from the motor torque characteristic, there is no braking power under normal operating conditions. However, as braking without load and feed torque must be taken into account, the braking power and braking energy are also calculated. In this case, the motor torque is obtained during the deceleration phase: M mot v = -( J mot + J gearbox ) load max i - J E load max gearbox i = -( 0.00665 + 0.00174) 189 7 - 0.0463 189 0.95 = -12.29 Nm 7 The max. braking power for the brake resistor is: Pbr W max = M mot v max i mot WR = -12.29 29 7 0.91 0.98 = -2.22 kW Braking energy and average braking power for one cycle: Wbr = 1 1 Pbr W max t v = - 2.22 01534 . = -017 . kWs 2 2 Pbr average = 344 Wbr 170 =- = -226.7 W T 0.75 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks Calculating the DC link currents when motoring The motor output is calculated from the motor torque and angular velocity: Pmot = M mot i 9000 8000 Motor power in W 7000 6000 5000 4000 3000 2000 1000 0 0 0.1 0.2 0.3 0.4 Time in s 0.5 0.6 0.7 0.8 The following is valid for the max. DC link current: I DC link max = Pmot max mot INV U DC link = 8046 = 16.71 A 0.91 0.98 1.35 400 The RMS value of the DC link current is given by: I DC link RMS = 1 mot INV U DC link (P 2 mot i -1 2 + Pmot i + Pmot i Pmot i -1 ) T ti 3 for Pmot i -1 , Pmot i 0 A value of 7.42 A is obtained Siemens AG SIMOVERT MASTERDRIVES - Application Manual 345 3 Various special drive tasks 09.99 Mesh removal drive The mesh removal is a traversing drive which transports the meshes away. The transport is realized in 3 stages over a distance of 4.85 m. Initially, two movements are made, each with 0.5 m with intermediate pauses and then the remaining 3.85 m is traversed in one step. The drive traverses back to the initial position without any pause. Drive data Trolley weight mtrolley = 200 kg Load mass (mesh) mload = 120 kg Tension force FZ = 1900 N Friction force FR = 100 N Distance, total s = 4.85 m Max. traversing velocity vmax = 2.5 m/s Partial traversing distance s1 = 0.5 m/s Traversing time for the partial traversing distance ttot. 1 = 0.5 s Pause times tP = 0.375 s Pinion diameter D = 0.15152 m Clock cycle time T = 16 s 346 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks Traversing characteristics v forwards backwards v max v max 1 0.5 m 3.85 m 0.5 m 2 tP 1 1 t tP ttot. 1 tP 3 ttot. 1 4.85 m - v max T a a max t - a max A triangular velocity characteristic is used for the 2 partial traversing distances. Thus, the maximum velocity and the maximum acceleration are obtained as follows: v max 1 = a max = 2 s1 2 0.5 = = 2 m/ s t tot 1 0.5 2 v max 1 t tot 1 = 22 = 8 m / s2 0.5 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 347 3 Various special drive tasks 09.99 The same acceleration is assumed for the remaining distances. The missing times are obtained as follows: tb 2 = tv 2 = v max 2 a max 2.5 = 0.3125 s 8 = s2 - v max 2 t b 2 tk 2 = v max 2 tb 3 = t v 3 = v max 3 a max = s3 - v max 3 t b 3 tk 3 = v max 3 = 385 . - 2.5 0.3125 s = 12275 . 2.5 2.5 = 0.3125 s 8 = 4.85 - 2.5 0.3125 s = 16275 . 2.5 Load force characteristics F load forwards backwards F +F z R t -F R T 348 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks Load torque The load torque is given by: M load forwards = J load forwards load + Fload D 2 M load backwards = J load backwards load + Fload D 2 with load = a J load 2 D forwards D = ( mtrolley + mload ) ( ) 2 2 D J load backwards = mtrolley ( ) 2 2 a from characteristic a(t) Fload from the characteristic Fload(t) 350 300 250 Load torque in Nm 200 150 100 50 0 0 2 4 6 8 10 12 14 16 -50 -100 -150 Time in s Siemens AG SIMOVERT MASTERDRIVES - Application Manual 349 3 Various special drive tasks 09.99 Selecting the motor Selected motor: 1FK6103-8AF71. with gearbox i=7 (gearbox=0.95), nn=3000 RPM, Jmot=0.01215 kgm2, mot=0.94, Jgearbox=0.00185 kgm2, kT0=1.51 Nm/A, M0=36 Nm, I0=23.8 A, Mmax=102 Nm, Imax=78 A Maximum motor speed: nmot max = v max 60 i 2.5 60 7 = = 2206 RPM . D 015152 Motor torque The motor torque is given by: M mot forwards = J mot load i + J gearbox load i + M load M mot backwards = J mot load i + J gearbox load forwards 1 i VZ gearbox VZ gearbox i + M load backwards i with VZ = sign( M load forwards ,backwards ) 70 60 50 Motor torque in Nm 40 30 20 10 0 0 2 4 6 8 10 12 14 16 -10 -20 -30 Time in s 350 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks 120 M mot, M per. in Nm 100 80 60 40 20 0 0 500 1000 1500 2000 2500 3000 Motor speed in RPM The motor torque lies below the limiting characteristic. Thus, the motor is suitable as far as the dynamic limits are concerned. Thermally checking the motor The RMS motor torque is calculated from the motor torque characteristic: M RMS = M 2 mot i ti T = 16.75 Nm The average motor speed is given by: naverage = nmot i-1 + nmot i 2 T ti nmot max 1 = 2 t b 1 4 + nmot max ( t b 2 2 + t k 2 + t k 3 ) T 2 60 7 0.25 4 + 2206 ( 0.3125 2 + 12275 . + 16275 . ) 2 015152 . = = 534.9 RPM 16 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 351 3 Various special drive tasks 09.99 40 Motor torque in Nm 35 30 25 M per. S1 M RMS / n average 20 15 10 5 0 0 500 1000 1500 2000 2500 3000 Motor speed in RPM The effective motor torque lies, at naverage, below the MS1 characteristic. This means that the motor is also suitable as far as the thermal limits are concerned. Selecting the inverter 43.6 A is obtained for the peak current. The RMS value is 11.44 A. Selected SIMOVERT MASTERDRIVES MOTION CONTROL inverter: 6SE7023-0TP50 (Compact Plus type of construction) Pinv n=15 kW; Iinv n=34 A; IDC link n=40.5 A; Iinv max=54.4 A (160 % overload capability) 352 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks 45.00 40.00 35.00 Motor current in A 30.00 25.00 20.00 15.00 10.00 5.00 0.00 0 2 4 6 8 Time in s 10 12 14 16 Calculating the braking power and the braking energy The motor output is obtained from the motor torque and the traversing velocity: Pmot = M mot i 2v D 15000 Motor power in W 10000 5000 0 0 2 4 6 8 10 12 14 16 -5000 -10000 Time in s Siemens AG SIMOVERT MASTERDRIVES - Application Manual 353 3 Various special drive tasks 09.99 The max. braking power for the braking resistor is obtained through the maximum negative motor output (=motor braking power): Pbr W max = Pbr mot max mot inv = -5952.9 0.94 0.98 = -5483.8 W The braking energy for one cycle is obtained through the negative surface area under the motor output characteristic 1 Wbr = mot inv Pbr mot max i t vi = -2078 Ws 2 The average braking power for one cycle is Pbr average = Wbr 2078 =- = -129.9 W T 16 Calculating the DC link currents when motoring The following is valid for the max. DC link current: I DC link max = Pmot max mot inv U DC link = 14390 = 28.93 A 0.94 0.98 1.35 400 The DC link RMS current is obtained as follows: I DC link RMS 1 = mot inv U DC link (P 2 mot i -1 2 + Pmot i + Pmot i Pmot i -1 ) T ti 3 for Pmot i -1 , Pmot i 0 A value of 4.75 A is obtained 354 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks Selecting the rectifier unit The rectifier unit is selected according to the peak DC link current and the DC link RMS current. I DC link max I DC link EE max I DC link RMS I DC link EE n Further, the rated DC link current of the rectifier unit should be at least 30% of the total of the rated DC link currents of the connected inverters. 0.3 I DC link INV n I DC link EE n For the 15 kW rectifier unit Compact Plus, the sum of the rated DC link currents of the connected inverters may not be greater than 80 A due to the resistor pre-charging. As the four drives can operate simultaneously, the sum of the peak values of the individual inverters is used to estimate IDC link max (this means that the calculation is on the safe side). I DC link max I DC link INV max = 14.44 + 48.02 + 2 16.71 + 28.93 = 124.8 A The RMS values of the individual inverters are added to determine IDC link RMS. In principle, the RMS values of the total DC link currents must be generated, but to save time this has not been done and the error neglected. I DC link RMS I DC link INV RMS = 3.76 + 16.33 + 2 7.42 + 4.75 = 39.68 A When the rated DC link currents are added for the individual inverters, the following is obtained: I DC link INV n = 30.4 + 44.6 + 2 20.8 + 40.5 = 157.1 A Siemens AG SIMOVERT MASTERDRIVES - Application Manual 355 3 Various special drive tasks 09.99 Selected SIMOVERT MASTERDRIVES MOTION CONTROL rectifier unit 6SE7031-2EP85-0AA0 (Compact Plus type) Pn=50 kW; IDC link n=120 A; IDC link max=192 A (160 % overload capability) Selecting the brake resistor When selecting the brake resistor, it is assumed that the four drives must brake simultaneously. Thus, the sum of the individual peak braking powers is formed (in this case you will always be on the safe side). Pbr max Pbr INV max = 2370 + 5980 + 2 2220 + 5484 = 18274 W The average braking powers of the individual inverters are also added. Pbr average Pbr INV average = 62.94 + 800 + 2 226.7 + 129.9 = 1447 W With the conditions Pbr max 15 . P20 Pbr average P20 4.5 a braking resistor with P20=20 kW (6SE7023-2ES87-2DC0) is obtained. The chopper for this inverter is, for Compact Plus, integrated in the rectifier unit. 356 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3.16 3 Various special drive tasks Drive for a pantograph For the systems which have been implemented up until now, the pantograph is brought into the up (operating position) using a spring which is tensioned in the quiescent (down) position. This spring also provides the contact force between the pantograph bar and the overhead wire. The disadvantage with this arrangement is that the spring is continuously tensioned in the quiescent position, which means that the components are subject to increased mechanical stressing. The pantograph is brought back into the quiescent state, and the spring is either pneumatically tensioned or tensioned using an auxiliary electric drive. Overhead wire Operating position (up position) Bar Quiescent position Lower joints Principle mode of operation When the spring is replaced by an electric drive, the drive must fulfill the following requirements: * The pantograph must be retracted from the overhead wire within approx. 1 s * After the bar comes into contact with the overhead wire, it must be pressed onto the overhead wire with a constant force * When the overhead wire height changes, the drive must appropriately correct the pantograph so that the contact force is, as far as possible, constant. Siemens AG SIMOVERT MASTERDRIVES - Application Manual 357 3 Various special drive tasks 09.99 Initially, the load torque is investigated at one of the two lower joints of the pantograph as a function of angle . The motor torque can, for example, be applied through a gearbox, or an additional lever arm with rack and pinion. Positioning can be implemented using a suitable traversing cam. After the pantograph bar makes contact with the overhead wire, the motor must be closed-loop torque controlled using a torque limiting function as a function of the height in order to guarantee a constant contact force. Pantograph data Weight of the upper pantograph assembly (left and righthand section) mo = 64 kg Weight of the lower pantograph assembly (left and righthand section) mu = 80 kg Weight of the bar mW = 60 kg Length of the upper bars lo = 1.751 m Length of the lower bars lu = 1.25 m Contact force FA = 160 Nm Height in the lower position yu = 0.416 m Height in the upper position yo = 2.481 m Distance between the lower joints 2a =1m Positioning time ttot =1s Angular relationships 3 lo 2 y lu 1 a 358 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks The height y of the pantograph is a function of angle : y = l u sin + lo2 - ( a + lu cos ) 2 Equation 1 The following is valid for the angle: lo2 - (a + l u cos ) 2 = arcsin , lo = - 2 Using equation 1, and the given heights yu and yo, the associated angle u and o can be iteratively determined. The following is obtained: u = 0.15 rad (lower quiescent position) o = 1.085 rad (working (up) position) 3 2.5 y in m 2 y yu yo 1.5 1 0.5 0 0 0.2 0.4 0.6 0.8 1 1.2 1.4 Anlge phi in rad Diagram to determine the initial values for iteration Movement Starting from the lower quiescent position, the pantograph should be brought into its final position within 1 s. It must also be retracted to its final position in the lower quiescent position also within Siemens AG SIMOVERT MASTERDRIVES - Application Manual 359 3 Various special drive tasks 09.99 1 s. The angular velocity at the lower joint 1 is specified as symmetrical triangular traversing characteristic. Thus, the following characteristic is obtained for the angular velocity and the angular acceleration. load load max t t tot. tp - load max load load max t - load max The following is obtained for the max. angular velocity, max. angular acceleration and the max. load speed: load max = load max = nload max = 360 2 2 ( o - u ) 2 (1.085 - 0,15) = = = 1.87 s -1 t ges t ges 1 2 load max t ges = 2 1.87 = 3.74 s -2 1 load max 60 1.87 60 = = 17.86 min -1 2 2 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks The angular acceleration load, the angular velocity load and the angle can now be represented as a function of time. Range: 0 t t b (acceleration upwards) load = load max load = load max t 1 = u + load max t 2 2 Range: tb t ttotal (acceleration downwards) load = - load max load = - load max (t - ttotal ) 1 = u + - load max (t - ttotal ) 2 2 Range: ttotal + t p t ttotal + t p + tb (acceleration upwards) load = - load max load = - load max (t - ttotal - t p ) 1 = u + - load max (t - ttotal - t p ) 2 2 Range: ttotal + t p + tb t ttotal + t p + ttotal (acceleration downwards) load = - load max load = load max (t - 2 ttotal - t p ) 1 = u + load max (t - 2 ttotal - t p ) 2 2 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 361 3 Various special drive tasks 09.99 Calculating the load torque when raising and lowering the pantograph When raising and lowering the pantograph, the load torque is calculated, without taking into account the contact force. Torque as result of the forces due to the various weights mo mW + 4 2 Fo FW + 4 2 3 lo mo 4 2 Fo 4 lu Fu 2 MG 2 1 The forces associated with only one half of the pantograph are investigated due to the symmetry. Half of the weight of the upper assembly Fo/2 is distributed, extending from the center of gravity, 50% to point 2 and 3 in order to simplify the calculation. The torque associated with point 1 does not change. Half of the weight of the lower assembly Fu/2 acts at the center of gravity. Only half the weight of the counterpoise FW /2 acts at point 3. The force Fo/4+FW /2, acting vertically at point 3, is broken down into a bar stress for bar l o, and then into a normal component force for bar lu. 362 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks FSt G Fo + FW 4 2 lo FSt G FN G MG 2 lu F 4 Fu 2 1 The following is obtained for the bar stress and the normal component of the force: FSt G Fo FW m m + ( o + W )g 2 = 4 2 = 4 cos cos FN G = FSt G cos with = - - 2 Thus, the torque as result of the various weights is given by: M G Fu lu F = cos + o l u cos + FN G lu 2 2 2 4 = mu l m g u cos + o g lu cos + FN G lu 2 2 4 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 363 3 Various special drive tasks 09.99 Torque due to the acceleration of the various masses y, v y , a y mo mW + 4 2 3 Fy 2 lo mo 4 2 lu mu 2 Mb 2 1 Also here, due to the symmetry, only the forces associated with half of the pantograph are investigated. The force Fy/2 is obtained from the acceleration in the y axis: Fy 2 =( mo mW + )ay 4 2 The velocity vy and acceleration ay is obtained using equation 1 by differentiation. vy = (a + lu cos ) l u load sin dy d = l u load cos + d dt y - lu sin ay = dv y d T = d dt y - l u sin with 2 2 T = a lu ( load cos + load sin ) + lu (- y load sin + cos ( y load + 2 v y load )) - v y2 364 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks Also here the force Fy/2, acting vertically in point 3, is broken down into a bar force regarding bar lo and then into a force to the vertical regarding bar lu. mo 4 mu 2 FN b Mb 2 lu 1 The bar stress and the normal component of the force as result of the acceleration are given by: Fy FSt b = 2 cos FN b = FSt b cos Thus, the torque due to acceleration is given by: Mb = J 1 load + FN b l u 2 In this case, J1 is the moment of inertia of bar lu with weight mu/2 and the weight mo/4, acting at point 2, referred to point 1. m 1 m J 1 = u lu2 + o l u2 3 2 4 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 365 3 Various special drive tasks 09.99 The load torque at point 1, when taking into account both halves of the pantograph is given by: M load = 2 ( MG Mb + ) 2 2 Using the numerical values, the load torque can be calculated, for example, using an Excel program. 7000 Upwards Downwards 6000 Load torque in Nm 5000 4000 3000 2000 1000 0 0 0.5 1 1.5 2 2.5 3 3.5 4 -1000 Time in s Load torque when the pantograph is being raised and lowered (pantograph moving up and down) Calculating the steady-state load torque When calculating the steady-state load torque during operation, the contact force FA is taken into account. The following is obtained: M load stat = 2 ( 366 mu l m g u cos + o g l u cos + FN stat lu ) 2 2 4 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks with mo mW F + ) g + A 2 2 cos = 4 cos ( FN stat 4000 3500 M load stat in Nm 3000 2500 2000 1500 1000 500 0 0 0.2 0.4 0.6 0.8 1 1.2 Angle phi in rad Steady-state load torque with contact force as a function of angle Selecting the motor When selecting the motor a simplification is made in so much that it is assumed that the motor torque is converted to the load torque linearly through a gearbox. Selected motor: 1PA6105-4HF. with gearbox i=110 (gear=0.9), Mn=44 Nm, nn=17500 RPM, Mstall=114.4 Nm, In=17.5 A, I=9.3 A, JMot=0.029 kgm2, Jgear=0.02 kgm2, Mot=0.875 Maximum motor speed: nmot max = i nload max = 110 17.86 = 1964.3 RPM Siemens AG SIMOVERT MASTERDRIVES - Application Manual 367 3 Various special drive tasks 09.99 Motor torque The following is valid for the motor torque: M mot up = J mot load i + J gear load i + M load M mot down = J mot load i + J gear load M mot stat = M load stat 1 i VZ gear VZ i + M load gear i 1 i with VZ = sign( M load ) 100 Upwards Downwards 80 Motor torque in Nm 60 40 20 0 0 0.5 1 1.5 2 2.5 3 3.5 4 -20 -40 Time in s Motor torque when the pantograph is being raised and lowered (pantograph is moved up and down) 368 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks 40 35 M Mot stat in Nm 30 25 M Mot G M Mot stat 20 15 10 5 0 0 0.2 0.4 0.6 0.8 1 1.2 Angle phi in rad Steady-state motor torque with contact force (Mmot stat) and without contact force (Mmot G) as a function of angle The highest motor torque in dynamic operation is required when the pantograph is being raised (moved upwards). This torque is 81.13 Nm. The motor torque when the pantograph is being lowered (moved downwards) is less due to the efficiency of the gear. In order to achieve a constant contact force in steady-state operation, the motor torque limit must be entered as a function of angle according to the characteristic Mmot stat. This can be realized, for example, by sensing the angle using an absolute value encoder and entering the torque limit via a characteristic block. In order to compensate overhead wire height changes more quickly, a torque increase can be used which is dependent on the differentiated angle . Mmot stat is greater than torque Mmot G, caused by the weight, by the component of the contact force FA. The maximum steady-state motor torque is 33.55 Nm. However, this value is not reached, as the overhead wire height only varies by 0.5 m. Thus, using equation 1, an angle of 0.8 rad and a steady-state motor torque of approximately 25 Nm is obtained. Thus, the motor is thermally suitable, as the rated torque is 44 Nm. Siemens AG SIMOVERT MASTERDRIVES - Application Manual 369 3 Various special drive tasks 09.99 100 90 M mot, M perm. in Nm 80 70 60 M Mot dyn M perm. 50 40 30 20 10 0 0 500 1000 1500 2000 2500 Motor speed in 1/min The dynamic motor torque lies below the limiting curve. Thus, the motor is also suitable regarding the dynamic limits. Dimensioning the brake resistor The motor output is obtained from the motor torque and the angular velocity as follows: Pmot = M mot i load 370 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks 15000 Upwards Downwards Motor power in W 10000 5000 0 0 0.5 1 1.5 2 2.5 3 3.5 4 -5000 -10000 -15000 Time in s Motor output while the pantograph is being raised and lowered (pantograph moving up and down) The max. braking power for the brake resistor is obtained from the max. negative motor output (=motor braking power): Pbr W max = Pbr mot max mot INV = -10.7 0.875 0.98 = -9.17 kW As a result of the condition Pbr W max 1.5 P20 a braking unit with minimum P20=10 kW must be used. The braking energy when rasing and lowering the pantograph can be calculated using the area under the negative motor power curve. Wbr = Pmot mot INV dt mot INV for Pmot 0 Pmot i -1 + Pmot i 2 (t i - t i -1 ) Siemens AG SIMOVERT MASTERDRIVES - Application Manual for Pmot i -1 , Pmot i 0 371 3 Various special drive tasks 09.99 The calculation results in: Wbr = -3.58 kWs The following must be valid for a braking unit with internal brake resistor: Wbr P20 T 36 or T Wbr 36 3.58 36 = = 12.9 s P20 10 (at P20=10 kW) When a 10 kW braking unit with internal brake resistor is used, the cycle time for upwards and downwards motion must therefore be a minimum of 12.9 s. Otherwise, an external brake resistor must be used or possibly a larger braking unit. Selecting the drive converter The drive converter is selected according to the max. motor current when accelerating and the max. motor current under steady-state operating conditions. The motor current is calculated from the motor torque as follows: I mot I mot n ( I mag 2 1 I mag 2 M mot 2 ) (1 - ( ) ) kn 2 + ( ) 2 I mot n M mot n I mot n kn kn = 1 kn = nmot nmot n for n mot n mot n constant flux range for n mot > n mot n field weakening Thus, the following is obtained: I mot b max = 31.2 A max. motor current when accelerating with 81.13 Nm I mot stat max = 12.6 A max. motor current under steady-state operating conditions with 25 Nm 8 % is added for Imot b max for saturation effects. 372 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks Selected drive converter SIMOVERT MASTERDRIVES VECTOR CONTROL: 6SE7022-6EC61 PU n=11 kW; IU n=25.5 A; IU max=34.8 A (136 % overload capability) with 10 kW braking unit (6SE7021-6ES87-2DA0) 35 Upwards Downwards 30 Motor current in A 25 20 15 10 5 0 0 0.5 1 1.5 2 2.5 3 3.5 4 Time in s Motor current when the pantograph is being raised and lowered. Siemens AG SIMOVERT MASTERDRIVES - Application Manual 373 3 Various special drive tasks 09.99 Estimating the accuracy of the contact force If the steady-state torque characteristics at an operating point are investigated, then the most unfavorable case at the lowest required motor torque is obtained, i. e. at the maximum height: M mot stat = 17.8 Nm steady-state motor torque M mot G = 15.7 Nm motor torque component for the forces due to the weight and therefore M mot A = M mot stat - M mot G = 2.1 Nm ~ FA motor torque component for the contact force If a 1% torque accuracy is used for the drive, then the following is obtained: M mot = Mn 44 1% = = 0.44 Nm 100 100% Thus, the accuracy of the contact force is given by: M mot A M mot A = FA M mot 0.44 = = 100% = 21% 2 .1 FA MA Due to the fact that the motor has been dimensioned according to the dynamic requirements, the required steady-state motor torque is relatively small with respect to the rated motor torque and because the component responsible for the contact force is only approximately 12 % of the steady-state motor torque, a larger error is obtained regarding the accuracy of the contact force. Better results would be achieved if the component of the weight could be kept lower, e. g. using an equalization weight, or a spring, which would compensate the weight component in the operating range. 374 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks Checking the calculation using the potential energy The calculation can be checked using the potential energy to move the masses from the quiescent position into the operating position. To realize this, the center of gravity travel of weights mo, mu and mW is calculated. h mo = yo + lu sin o y u + lu sin u 2.481 + 1.25 sin 1.085 0.416 + 1.25 sin 0.15 - = - = 1.492 m 2 2 2 2 h mu = lu sin o lu sin u 1.25 sin 1.085 1.25 sin 0.15 - = - = 0.459 m 2 2 2 2 h mW = yo - y u = 2.481 - 0.416 = 2.065 m Thus, the following is obtained for the potential energy: W pot = mo g h mo + mu g h mu + mW g h mW = (64 1.492 + 80 0.459 + 60 2.065) 9.81 = 2512 Ws The following is obtained for the average power: Paverage = W pot t total = 2512 = 2512 W 1 On the other hand, the energy for the upwards motion is calculated as follows: Wup = t total Pload dt = 0 t total M load load dt 0 M load i -1 load i -1 + M load i load i 2 (ti - ti -1 ) When this relationship is evaluated, for appropriately smaller steps, the same value is determined as for W pot. The kinetic energy, required when accelerating, is completely fed back when decelerating. Siemens AG SIMOVERT MASTERDRIVES - Application Manual 375 3 Various special drive tasks 09.99 8000 7000 6000 5000 P in W 4000 P load P average 3000 2000 1000 0 0 0.2 0.4 0.6 0.8 1 1.2 -1000 -2000 Time in s Load power Pload and average power Paverage when moving upwards (pantograph being raised) 376 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks Drive for a foil feed with sin2 rounding-off 3.17 The drive will be investigated, simplified as a pure flywheel drive. An 1FK6 motor is to be used. In 2 order to avoid torque steps, the speed setpoint is rounded-off according to a sin characteristic (also refer under 4.6). Drive data Load moment of inertia Jload = 0.75 kgm2 Gearbox ratio i = 15.57 Gearbox moment of inertia JG = 0.000151 Gearbox efficiency G = 0.95 Max. motor speed nMot max = 3000 RPM Accelerating time tb = 70 ms Deceleration time tv = 70 ms Constant speed time tk = 70 ms No-load time tp = 0.8 s The maximum angular velocity of the load is given by: load max = 2 nmot max 60 i = 2 3000 = 20.1772 s -1 60 15.57 For a sin2 rounding-off function, the maximum angular acceleration of the load is given by: load max = load max 20.1772 = = 452.8 s - 2 tb 2 0.07 2 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 377 3 Various special drive tasks 09.99 Traversing curve load load max t tb tk tv tp load load max t - load max The traversing curve is symmetrical due to tb=tv. 378 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks The angular velocity load and the angular acceleration load can be represented as a function of the time as follows: Range: 0 t t b (acceleration) load = load max sin 2 ( load = load max sin( t ) 2 tb t ) tb Range: t b t t b + t k (constant speed) load = load max load = 0 Range: t b + t k t t b + t k + t v load = load max cos 2 ( load = - load max sin( (deceleration) (t - tb - tk ) ) 2 tv (t - tb - tk ) ) tv Load torque The following is valid for the load torque: M load = J load load The maximum load torque is given by: M load max = J load load max = 0.75 452.8 = 339.6 Nm Siemens AG SIMOVERT MASTERDRIVES - Application Manual 379 3 Various special drive tasks 09.99 Selecting the motor Selected motor: 1FK6063-6AF71, nn=3000 RPM, JMot=0.00161 kgm2, Mot=0.89, kT0=1.33 Nm/A, M0=11 Nm, I0=8.3 A, Mmax=37 Nm, Imax=28 A Motor torque The following is valid for the motor torque: M mot b = (( J mot + J G ) i + J load 1 t ) load max sin( ) i G tb M mot v = -(( J mot + J G ) i + J load (acceleration, motoring) G (t - tb - tk ) ) load max sin( ) i tv (deceleration, generating) 40 30 Motor torque in Nm 20 10 0 0 0.2 0.4 0.6 0.8 1 1.2 -10 -20 -30 -40 Time in s The highest motor torque is achieved when accelerating, with: M mot b max = (( J mot + J G ) i + J load 1 ) load max i G = ((0.00161 + 0.000151) 15.57 + 0.75 380 1 ) 452.8 = 35.37 Nm 15.57 0.95 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks The max. motor torque when decelerating is lower as a result of the gearbox efficiency: M mot v max = -(( J mot + J G ) i + J load G ) load max i = -((0.00161 + 0.000151) 15.57 + 0.75 0.95 ) 452.8 = -33.14 Nm 15.57 40 35 M mot, M perm. in Nm 30 25 20 15 10 5 0 0 500 1000 1500 2000 2500 3000 3500 Motor speed in RPM The motor torque lies below the limiting curve. Thus, the motor is suitable as far as the dynamic limits are concerned. Thermally checking the motor The RMS motor torque is given by: tb M RMS = M 0 2 mot b dt + tb + t k + tv M 2 mot v dt tb +t k tb + t k + t v + t p Siemens AG SIMOVERT MASTERDRIVES - Application Manual 381 3 Various special drive tasks 09.99 After inserting, the following is obtained: tb M M RMS = 2 mot b max 0 t sin ( ) dt + tb 2 tb +t k + tv 2 2 M mot v max sin ( tb + t k (t - t b - t k ) ) dt tv tb + t k + tv + t p tb tv 0.07 0.07 2 + M mot 35.37 2 + 33.14 2 v max 2 2 = 2 2 = 9.02 Nm tb + t k + tv + t p 0.07 + 0.07 + 0.07 + 0.8 2 M mot b max = The average motor speed is given by, with nmot ~ load: tb + t k +tv n average = = n t n mot max ( sin ( ) dt + 2 tb 0 tb mot dt 2 0 tb + t k + t v + t p tb = tb + t k tb dt + tb + t k + t v tb + t k cos 2 ( (t - t b - t k ) ) dt ) 2 tv tb + tk + tv + t p tv 0.07 0.07 ) 3000 ( + 0.07 + ) 2 2 = 415.8 RPM 2 2 = tb + t k + tv + t p 0.07 + 0.07 + 0.07 + 0.8 nmot max ( + tk + 12 10 Torque in Nm 8 6 M perm. S1 M RMS / n average 4 2 0 0 500 1000 1500 2000 2500 3000 3500 Motor speed in RPM 382 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks The RMS motor torque is, at naverage below the MS1 characteristic. Thus, the motor is suitable as far as the thermal limits are concerned. Dimensioning the brake resistor The motor output is obtained from the motor torque and the angular velocity as follows: Pmot = M mot i load 8000 6000 Motor power in W 4000 2000 0 0 0.2 0.4 0.6 0.8 1 1.2 -2000 -4000 -6000 -8000 Time in s The following is valid when generating: Pmot v = - M mot v max sin( (t - t b - t k ) (t - t b - t k ) ) i load max cos 2 ( ) tv 2 tv The max. braking power of the brake resistor is obtained from the max. motor output when generating: Pbr W max = Pmot v max mot INV Siemens AG SIMOVERT MASTERDRIVES - Application Manual 383 3 Various special drive tasks 09.99 If the following is used (t - t b - t k ) = 2 tv then the maximum of Pmot v is obtained as follows, with dPmot v d = 0 = - M mot v max i load max (2 cos(2 ) cos 2 ( ) - 2 sin(2 ) cos( ) sin( )) for 1 = arcsin( ) = 0.5236 rad 2 By inserting , the maximum braking power for the brake resistor is given by: Pbr W max = - M mot v max i load max mot INV sin(2 0.5236) cos 2 (0.5236) = -33.14 15.57 20.1772 0.89 0.98 sin( 2 0.5236) cos 2 (0.5236) = -5898 W The braking energy for a cycle is calculated as follows: Wbr = tb + t k + t v P mot v mot INV dt tb + t k = - M mot v max i load max mot INV tb +t k +t v sin( tb +t k = - M mot v max i load max mot INV (t - t b - t k ) (t - t b - t k ) ) cos 2 ( ) dt tv 2 tv tv 0.07 = -33.14 15.57 20.1772 0.89 0.98 = -202.3 Ws The following is valid for the brake resistor: Pbr W max = 5898 W 1.5 P20 P Wbr 202.3 = = 200.3 W 20 0.07 + 0.07 + 0.07 + 0.8 4 .5 T (ext. brake resistor) Thus, when selecting a drive converter, type of construction Compact Plus, an external 5 kW brake resistor is required (6SE7018-0ES87-2DC0). The chopper is integrated into the drive converter. 384 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 3 Various special drive tasks Selecting the drive converter The drive converter is selected according to the peak- and the RMS value of the motor current. Refer to example 3.13 for a precise calculation of the current. The peak current is 27.1 A. The RMS value is 6.86 A. In this case, 300 % overload capability can be utilized for type of construction Compact Plus, as the overload time is less than 250 ms, after the overload current drops back down again to below 0.91 x of the rated current, and the recovery time is greater than 750 ms. An approximate calculation of the motor current results in: I mot max I RMS M mot max kT0 = 35.37 = 26.6 A 1.33 M RMS 9.02 = = 6.78 A kT0 1.33 Selected SIMOVERT MASTERDRIVES MOTION CONTROL drive converter: 6SE7021-0EP50 (Compact Plus type of construction) PINV n=4 kW; IU n=10 A; IU max=30 A (300 % overload capability) A 7.5 kW drive converter could be used without utilizing the 300 % overload capability IU max=32.8 A (160 % overload capability). 30.00 Motor current in A 25.00 20.00 15.00 10.00 5.00 0.00 0 0.2 0.4 0.6 0.8 1 1.2 Time in s Siemens AG SIMOVERT MASTERDRIVES - Application Manual 385 3 Various special drive tasks 386 09.99 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 4 Information for specific applications 4 Information for specific applications 4.1 Accelerating time for loads with square-law torque characteristics 4.1.1 General information If the load torque is specified in the following form for fan and pump drives M load = M load max ( n n max )2 at constant motor torque (e.g. when accelerating along the current limit in the constant-flux range), the accelerating time can be calculated using the following formula: ta = n max J ln 60 M motor M load max M motor +1 M load max accelerating time from 0 to nmax M motor -1 M load max [s] J in kgm2 (moment of inertia of the motor, load and coupling), M in Nm, n in RPM M M motor Accelerating torque M load max M load n n max Example of the motor- and load torque when starting Siemens AG SIMOVERT MASTERDRIVES - Application Manual 387 4 Information for specific applications 09.99 The difference between the motor- and load torque acts as accelerating torque. After acceleration to nmax the motor torque and the load torque are the same. Acceleration is faster, the higher the ratio Mmotor / Mload max and the lower the total moment of inertia. If the load torque is not a square-load torque, or if the motor torque is not constant when accelerating, then the accelerating time must be numerically calculated using the following equation: J = 30 ta nmax 0 dn M motor ( n) - M load ( n ) By sub-dividing the function into m segments, the following approximation is obtained: ta J i=m 1 1 1 ( + ) ( n i - n i -1 ) 30 i =1 2 M motor i -1 - M load i -1 M motor i - M load i Mmotor and Mload must be available in tabular form in order to evaluate this approximation. M M M motor motor i -1 M motor i M load i M load i -1 Mload n i -1 n i n max n Example for the motor torque- and load torque characteristics when sub-divided into segments 388 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 4 Information for specific applications 4.1.2 Accelerating time for a fan drive A fan is to accelerate up to nmax=4450 RPM. A gearbox with i = 1/3 is used. The power requirement at nmax is 67 kW. The load torque characteristic and moment of inertia of the fan are known. M load / M load max 1 0.8 0.6 0.4 0.2 0 0 0.2 0.4 0.6 0.8 1 n / n max Load torque characteristic of the fan The maximum load torque is: M load max = Pmax 9550 67 9550 = = 1438 . Nm n max 4450 Thus, the load torque characteristic is approximately obtained from the following: M load 143.8 Nm ( n 2 ) 4450 Moment of inertia of the fan: J fan = 25 kgm 2 A 4-pole 75 KW motor is selected (1LA6 280-4AA..). Mn=484 Nm, nn=1480 RPM, Imotor n=134 A, Jmotor=1.4 kgm2 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 389 4 Information for specific applications 09.99 Total moment of inertia referred to the motor shaft: J fan J = i2 + J motor = 32 25 + 1.4 = 225 + 14 . = 226.4 kgm 2 The load torque characteristic, referred to the motor shaft, is approximately obtained as follows: M load M load max i G ( n motor 2 ) i 4450 When assuming that the gearbox efficiency G is 1, the following is obtained: M load 3 143.8 Nm ( n motor 2 n ) = 431.4 Nm ( motor ) 2 4450 / 3 1483 The following drive converter is to be used: 6SE7031-8EF60 PV n=90 kW; IV n= IV continuous=186 A Closed-loop frequency control type The motor is to be accelerated with a current corresponding to the continuous drive converter current. Thus, the possible motor torque is approximated as follows: M max M n 2 2 I motor max - I n 2 2 I motor n - I n The following is obtained with I motor n = 134 A, I motor max = IV continuous = 186 A and I n = 0.35 I motor n : M max 186 2 - 0.352 134 2 484 = 694 Nm 134 2 - 0.352 134 2 The accelerating time from nmax=4450 RPM to nmotor max=1483 RPM is given by: ta 390 = n motor max J 60 M motor M load max ln M motor +1 M load max M motor -1 M load max Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 ta = 4 Information for specific applications 1483 226,4 ln 60 694 431,4 694 +1 4314 . = 68.6 s 694 -1 . 4314 As the accelerating time is greater than 60 s, the maximum drive converter current cannot be used for acceleration. Siemens AG SIMOVERT MASTERDRIVES - Application Manual 391 4 Information for specific applications 4.1.3 09.99 Accelerating time for a blower drive (using the field-weakening range) A blower is to be accelerated as fast as possible to nmax=3445 RPM using a drive converter. The load torque characteristic and the moment of inertia of the blower and coupling are known. M load / M load max 1 0.8 0.6 0.4 0.2 0 0 0.2 0.4 0.6 0.8 1 n / n max Load torque characteristic of the blower with M load max=116.7 Nm and nmax=3445 RPM The load torque characteristic is approximated as follows: M load 116.7 Nm ( n 2 ) 3445 Moment of inertia of the blower and coupling: J blower + J coupling = 9.275 kgm 2 + 01 . kgm 2 = 9.375 kgm 2 A 2-pole 90 KW motor is selected (1LA6 283-2AC..). Mn=289 Nm, nn=2970 RPM, Imotor n=152 A, Jmotor=0.92 kgm2 Total moment of inertia: J = J blower + J coupling + J motor = 9.375 kgm 2 + 0.92 kgm 2 = 10.295 kgm 2 392 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 4 Information for specific applications The motor should accelerate, in the constant-flux range, up to nn=2970 RPM with 200% of the rated torque along the current limit, and in the field-weakening range from nn=2970 RPM up to nmax=3445 RPM, with a current limit corresponding to the torque. Acceleration is sub-divided into two segments for the calculation. Range 1 (constant-flux range) The load torque is obtained as follows at the rated motor speed: M load 2970 = 116.7 ( 2970 2 ) = 86.74 Nm 3445 M / Nm 578 M motor = 2 M n M load 86.74 n / RPM 2970 Relationships/behavior in the constant-flux range In the constant-flux range, with Mmotor=2 Mn the accelerating time to nn=2970 RPM can be calculated as follows: ta 1 = 2970 J ln 60 2 M n M load 2970 2 Mn +1 M load 2970 2 Mn -1 M load 2970 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 393 4 Information for specific applications 09.99 2 289 +1 2970 10,295 86.74 ln ta 1 = = 5.84 s 60 2 289 86.74 2 289 -1 86.74 Range 2 (field-weakening range) For a constant motor current I motor , the motor torque ratios are approximated by: M field M = const fn f 2 I motor - I 2 n ( fn 2 ) f for f fn 2 - I 2 n I motor 2 As I motor I 2 n when starting, the expression can be simplified as follows: M field M = const f n nn f n for f fn (and n nn) Thus, the motor torque at nmax=3445 RPM is given by: M 3445 M 2970 nn n 2970 = 2 M n n = 2 289 = 498.3 Nm 3445 n max n max M / Nm 578 498.3 M motor M load 116.7 86.74 2970 3445 n / RPM Relationships in the field-weakening range The accelerating time in the field-weakening range can now be calculated as follows in one step: 394 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 4 Information for specific applications ta 2 = = J 1 ( 30 2 M motor 1 1 + ) (3445 - 2970) M motor 3445 - M load 3445 2970 - M load 2970 10.295 1 1 1 ( + ) 475 = 119 . s 30 2 578 - 86.74 498.3 - 116.7 Total accelerating time: t a = t a 1 + t a 2 = 584 . + 119 . = 7.03 s Comparing the accelerating time for direct online starting (without drive converter) For comparison purposes, the accelerating time to nn=2970 RPM is to be calculated when connected directly online to the 50 Hz line supply. The following torque characteristic is used as basis for the 90 KW motor with rotor class 10, MA/Mn=2 and Mstall/Mn=2.7. The starting characteristic is sub-divided into 15 segments. M / Nm 800 600 M motor 400 200 M load 0 0 1000 2000 3000 n / RPM Motor- and load torque when starting Siemens AG SIMOVERT MASTERDRIVES - Application Manual 395 4 Information for specific applications i n 0 09.99 0 Mload 8.17 Mmotor 578 ta 0 1 148 6 550 0.287 2 297 4.67 535 0.586 3 594 5.5 520 1.2 4 891 9 520 1.82 5 1188 13.9 530 2.45 6 1485 21.7 560 3.05 7 1782 31.2 600 3.63 8 2079 42.5 650 4.18 9 2376 55.5 710 4.69 10 2600 66.5 770 5.04 11 2673 70.3 780 5.15 12 2750 74.4 770 5.27 13 2800 77.1 740 5.35 14 2900 82.7 580 5.54 15 2970 86.7 289 5.8 Table to numerically calculate the starting (accelerating) time The accelerating time in the last column of the table, is calculated using the values entered for n, Mload and Mmotor by summing them according to the following approximation: ta = J i =15 1 1 1 ( + ) (ni - ni -1 ) 30 i =1 2 M motor i -1 - M load i -1 M motor i - M load i = 58 . s It is preferable to use a program such as Excel to evaluate this approximation formula. The direct online starting time is somewhat shorter, but the maximum speed of 3445 RPM is not reached. 396 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 4 Information for specific applications n / RPM 3000 2500 2000 1500 1000 500 0 0 1 2 3 4 5 6 t/s Speed characteristics for direct online starting up to nn=2970 RPM Selecting the drive converter The drive converter must be able to provide the necessary motor current to accelerate the 90 KW motor with 200% rated torque. The motor current, as a function of the motor torque is approximately given by: I motor ( M 2 2 2 2 ) ( I motor n - I n ) + I n Mn The following is obtained with I motor n = 152 A and I n = 0.25 I motor n for the 2-pole 90 KW motor: I motor 2 2 (152 2 - 0.252 152 2 ) + 0.252 152 2 = 296.8 A Drive converter selected: 6SE7032-6EG60 PV n=132 kW; IV n=260 A; IV max=355 A Closed-loop frequency control type Siemens AG SIMOVERT MASTERDRIVES - Application Manual 397 4 Information for specific applications 4.1.4 09.99 Re-accelerating time for a fan drive after a brief supply failure A fan drive is to accelerate to its previous (old) speed after a brief supply failure. The maximum power failure duration is 2 s. As a result of the relatively high drive moment of inertia, the kinetic buffering function is used to buffer the supply failure. Thus, the dead-times, which would occur if a restart-on-the-fly circuit was to be used, are eliminated: They include the motor de-excitation/ excitation times, the time to establish the power supply voltages and initialize the microprocessor system as well as the search time (the search time is eliminated when a tachometer/encoder is used). Drive data Rated motor output Pmotor n = 75 kW Rated motor current (at 500 V) Imotor n Rated motor torque Mmotor n = 241 Nm Maximum fan speed nmax = 2970 RPM Power requirement at nmax Pmax = 44.9 kW Moment of inertia of the fan Jfan = 4.75 kgm2 Moment of inertia of the motor Jmotor = 0.79 kgm2 = 102,4 A The maximum load torque is given by: M load max = Pmax 9550 44.9 9550 = = 144.4 Nm n max 2970 Thus, the load torque characteristic is approximated by: M load 144.4 Nm ( n 2 ) 2970 Total moment of inertia: J total = J fan + J motor = 4.75 + 0.79 = 554 . kgm2 To start off with, the motor speed must be defined at the instant that the line supply returns. The worst case condition is assumed with a maximum 2 s line supply failure and maximum speed when the power actually failed. The speed dip is the highest at maximum speed, as in this case, the highest load torque is effective. The kinetic buffering is active during the line supply failure. The motor speed is further decreased, in addition to the effect of the load torque, to cover the losses in the motor, inverter and in the 398 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 4 Information for specific applications power supply. The additional braking torque as a result of the losses can be neglected here, as the speed decrease during the line supply failure is low due to the relatively high moment of inertia and in the vicinity of nmax, the load torque component is the most significant. With this assumption, the following equation is obtained for the speed characteristic during line supply failure, starting from nmax, in the vicinity of nmax: nmin 1 nmax + 1 30 M load max 2 J total nmax tA The following is obtained for tA=2 s (line supply failure time) nmin 1 = 2544 RPM 1 30 144.4 2 + 2970 554 . 2970 2 The following drive converter is to be used: 6SE7031-3FG60 PV n=90 kW; IV n=128 A; IV max=174 A Closed-loop frequency control type The motor is to be accelerated with a current corresponding to the maximum drive converter current. The possible motor torque is approximated by: M max M n 2 2 I motor max - I n 2 2 I motor n - I n The following is obtained with I motor n = 102.4 A, I motor max = IV max = 174 A and I n = 0.28 I motor n : M max 241 174 2 - 0.28 2 102.4 2 = 420.7 Nm 102.4 2 - 0.28 2 102.4 2 The motor can therefore develop 420.7/241=175% rated torque when re-accelerating up to nmax. Siemens AG SIMOVERT MASTERDRIVES - Application Manual 399 4 Information for specific applications 09.99 The re-accelerating time from nmin=2544 RPM to nmax=2970 RPM is given by: = tH nmax J total (ln 60 M motor M load max M motor +1 M load max M motor -1 M load max 420.7 +1 2970 554 . 144 . 4 = (ln - ln 60 420.7 144.4 420.7 -1 144.4 - ln M motor n + min M load max nmax M motor n - min M load max nmax ) 420.7 2544 + 144.4 2970 ) = 0.84 s 420.7 2544 - 144.4 2970 Thus, the overall time from the start of the line supply failure up to when nmax is again reached is given by: t total = t A + t H = 2 + 0.84 = 2.84 s n / RPM 3000 2500 2000 1500 1000 500 0 0 0.5 1 1.5 2 2.5 3 t/s Speed characteristic for line supply failure (0 to 2 s) and re-starting (2 to 2.84 s) 400 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 4 Information for specific applications Taking into account the losses during kinetic buffering At low speeds, the braking torque due to the losses can no longer be neglected with respect to the load torque. Thus, the speed characteristic should now be determined during the kinetic buffer period by estimating the losses. Motor losses at the rated operating point: Pmotor n PV motor n = motor - Pmotor n = 75 - 75 = 4.45 kW 0.944 Under the assumption, that the motor is in a no-load condition during the kinetic buffering phase, and that the losses are then constant and approximately only half as high as they are at the rated operating point, then the following is obtained: PV no- load PV motor n 2 = 4.45 = 2.23 kW 2 If the losses in the inverter and in the power supply are neglected, then the additional braking torque is given by: M brake PV no- load 9550 n = PV no- load 9550 nmax n = M brake nmax max nmax n n n 2.23 9550 nmax = 7.2 Nm max n n 2970 The braking torque becomes higher at lower speeds. With this basis (assumption) for Mbrake, the speed characteristic during line supply failure, starting from nset, is given by: 2 a 2 - a nset + nset 2 nset - a 1 1 t c ( ln( )+ arctan( ) 2 6 (a + nset ) 3 a 3 a2 - a n + n 2 1 1 2 n - a - ln( )- arctan( )) 2 6 ( a + n) 3 a 3 with a = nmax 3 M brake nmax M load max c= nmax J total 2 30 3 M load max M brake nmax Siemens AG SIMOVERT MASTERDRIVES - Application Manual 401 4 Information for specific applications 09.99 This formula can no longer be used for speeds less than approximately 25 to 30% of the maximum speed, as the load torque characteristic approximation is too inaccurate (influence of the friction). As the formula cannot be changed-over for n, nmin must be found by inserting various values for n, with subsequent interpolation. Example for nset=750 RPM With a = 2970 3 7.2 = 1093 RPM 144.4 c= 2970 554 . 30 3 144.4 2 7.2 = 32.42 s and n=25 RPM, then t is shown in the following table i n t 0 750 0.00 1 725 0.38 2 700 0.76 3 675 1.13 4 650 1.50 5 625 1.86 6 600 2.22 The following value is obtained for nmin by linearly interpolating the values above and below tA=2 s (1.86 s and 2.22 s): nmin n6 + n5 - n6 625 - 600 (t 6 - t A ) = 600 + (2.22 - 2) = 615.3 RPM t 6 - t5 2.22 - 186 . The re-accelerating time from nmin=615.3 RPM to nset=750 RPM is then given by: tH = 402 nmax J total (ln 60 M motor M load max M motor n + set M load max nmax M motor n - set M load max nmax - ln M motor n + min M load max nmax M motor n - min M load max nmax ) Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 4 Information for specific applications 420.7 750 + 2970 5,54 144 . 4 2970 - ln = (ln 60 420.7 144.4 420.7 750 - 144.4 2970 tH 420.7 615.3 + 144.4 2970 ) = 019 . s 420.7 615.3 - 144.4 2970 Thus, the total time from the start of the line supply failure until nset is again reached is obtained from: t total = t A + t H = 2 + 019 . = 2.19 s n / RPM without taking into account the losses 750 taking into account the losses 500 250 t/s 0 0 0 .5 1 1 .5 2 2 .5 Speed characteristic for a line power failure (0 to 2 s) and re-acceleration (re-starting) (2 to 2.19 s) taking into account the losses during kinetic buffering (for comparison purposes, a curve is also shown where the losses were not taken into account) Siemens AG SIMOVERT MASTERDRIVES - Application Manual 403 4 Information for specific applications 09.99 Estimating nmin and tH nmin and tH can be estimated for short line supply failure times and high moments of inertia. Characteristics when coasting down With the gradient of the speed characteristic at the point n=nset dn =- dt n = nset M load max ( nset 2 n ) + M brake nmax max nmax nset Jtotal 30 then nmin is given by: nmin nset + t A dn dt n = nset If nset is close to nmax, then the approximation used earlier can be applied: nmin 1 30 M load max 1 + tA 2 nset Jtotal nmax Characteristics when re-accelerating With the gradient of the speed characteristic at the point n=nmin dn = dt n = nmin M motor - M load max ( Jtotal 30 nmin 2 ) nmax then tH is given by: tH 404 nset - nmin dn dt n=nmin Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 4 Information for specific applications Example for nset=750 RPM and tA=2 s (as previously calculated) 750 2 2970 144.4 ( ) + 7.2 dn 2970 750 = -65 RPM / s =- 554 . dt n = nset 30 nmin 750 + 2 ( -65) = 620 RPM 620 2 420.7 - 144.4 ( ) dn 2970 = 714.3 RPM / s = 554 . dt n = nmin 30 tH 750 - 620 = 0182 s . 714.3 Thus, the total time from the start of the line supply failure up to when nset is again reached is given by: ttotal = t A + t H = 2 + 0182 = 2.182 s . Siemens AG SIMOVERT MASTERDRIVES - Application Manual 405 4 Information for specific applications 4.1.5 09.99 Calculating the minimum braking time for a fan drive A fan drive is to be braked from nmax=1500 RPM to standstill using the existing brake resistor. It is assumed that the load has a square-law characteristic. Drive data Rated motor output (4 pole) Pmotor n = 250 kW Rated motor current Imotor n Rated motor torque Mmotor n = 1600 Nm Maximum fan speed nmax = 1500 RPM Power requirement at nmax Pmax = 198 kW Moment of inertia of the fan Jfan = 223 kgm2 Moment of inertia of the motor Jmotor = 3.6 kgm2 Motor efficiency motor = 0.962 Rated converter outputs PU n = 250 kW Brake resistor power using an ext. brake resistor P20 = 100 kW Peak braking power 1.5 P20 = 150 kW = 430 A The fan load torque helps to brake the drive - especially in the high speed range. To optimally utilize the brake resistor, it is therefore more favorable to brake with constant deceleration and not with constant braking torque (operation at the current limit). When decelerating with a constant braking torque, a higher peak braking power would be generated for the same braking time (refer to the comparison later on in the text). To optimally utilize the brake resistor, the peak braking power and the braking energy are important. Both values must remain below the permissible limits. Braking with constant deceleration Calculating the peak braking power for the brake resistor The motor torque when braking is given by: M motor br = M load + J 406 d dt Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 4 Information for specific applications Whereby M load M load max ( M load max = n nmax )2 Pmax 60 198000 60 = = 1260.5 Nm 2 nmax 2 1500 2 nmax d =- dt 60 tbr (constant deceleration) Thus, the motor torque is given by: M motor br = M load max ( n nmax )2 - J 2 nmax 60 t br and the motor output by: Pmotor br = M motor br = 2 n 60 2 n 2 J 2 nmax ( M load max ( ) - )n 60 nmax 60 t br The maximum motor output when braking, for n= 3 J 2 nmax 1 3 60 M load max t br is given by Pmotor br max = - 2 J 2 nmax J nmax 1 4050 10 M load max t br3 With a specified peak braking power Pbr W max for the brake resistor, and the following relationship Pmotor br max = Pbr W max motor Then the minimum permissible braking time can be calculated as follows: tbr min = J nmax 3 2 2 2 nmax motor 10 4050 2 M load max Pbr2 W max = (223 + 3.6) 1500 3 2 15002 0.962 2 = 17.52 s 10 40502 1260.5 150000 2 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 407 4 Information for specific applications 09.99 Calculating the braking energy when braking It must now be checked as to whether the permissible braking energy is exceeded. For a constant deceleration, the speed is given by: n = nmax (1 - t tbr ) and thus the motor output when braking Pmotor br = P 2 J 2 nmax 2 t t ( M load max nmax (1 - ) 3 - (1 - )) 60 t br 60 t br t br motor br t t 0 br t P motor br max Negative area corresponds to the braking energy Example for the motor output when braking with a square-law load torque characteristic The braking energy for the brake resistor is then given by: tbr Wbr = motor Pmotor br dt t0 = 2 motor 2 t br M load max nmax t J 2 n max t ( (1 - 0 ) 4 - (1 - 0 ) 2 ) 60 4 t br 2 60 t br t br Time t0 is obtained by setting Pmotor br to zero: t 0 = (1 - 408 J 2 n max ) t br 60 M load max t br Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 4 Information for specific applications If t0 is negative, a zero should be entered for t0 in the braking energy equation. In this case, even at the start of braking, the deceleration torque exceeds the load torque and t0 is given by: 226.6 2 1500 ) 17.52 = -4.72 s 60 1260.5 17.52 t 0 = (1 - t0 is therefore set to zero when calculating the braking energy, and the following is obtained: Wbr = 0.962 2 17.52 1260.5 1500 226.6 2 1500 2 ( - ) = -1855 kWs 60 4 2 60 17.52 The permissible braking energy for the external brake resistor is, for the 90 s interval: Wbr perm. = Pbr W time 90 s = 25 90 = 2250 kWs Thus, the second condition is also fulfilled for intervals, which are not less than T= Wbr 1855 90 s = 90 = 74.2 s Wbr perm. 2250 Diagrams for motor output and motor torque when braking with constant deceleration Motor breaking power for tbr=17.52 s 0 Motor braking power in kw 0 2 4 6 8 10 12 14 16 18 -20 -40 -60 -80 -100 -120 -140 -160 Time in s Siemens AG SIMOVERT MASTERDRIVES - Application Manual 409 4 Information for specific applications 09.99 Motor torque for tbr=17.52 s 0 0 2 4 6 8 10 12 14 16 18 Motor torque in Nm -500 -1000 -1500 -2000 -2500 Time in s Checking the drive converter dimensioning The maximum motor torque when braking occurs at n = 0, and is given by: M motor br max = - J 2 n max 226.6 2 1500 =- = -2032 Nm 60 t br 60 17.52 For the highest required motor torque when braking, an approximate motor current is given by: I motor max ( M motor max M motor n 2 2 2 ) 2 ( I motor n - I n ) + I n With I n = 0.31 I motor n (assumption), the following is obtained: I motor max ( 2032 2 ) (430 2 - 0.312 430 2 ) + 0.312 430 2 = 536 A 1600 Drive converter selected: 6SE7035-1EK60 PV n=250 kW; IV n=510 A; IV max=694 A This drive converter is adequately dimensioned for braking. 410 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 4 Information for specific applications Comparison, braking with a constant motor torque The maximum braking power is then always at n=nmax. The following is obtained for the required motor torque using the specified peak rating of the brake resistor : M motor br = - Pbr W max 60 2 nmax motor =- 150000 60 = -993 Nm 2 1500 0.962 Thus, the minimum braking time is obtained as follows: tbr min = = J 2 nmax 60 M load max M motor br arctan M load max M motor br 226.6 2 1500 1260.5 arctan = 26.9 s 993 60 1260.5 993 Thus, with the same peak braking power, the braking time is longer, than when braking with constant motor torque. The motor output during braking is given by: Pmotor br = M motor br 2 n 60 with n= n max M load max tan(arctan( M load max M motor br )- 60 M load max M motor br J 2 n max t) M motor br In this case, the braking energy can only be numerically calculated. When sub-dividing into m segments, for example, the following approximation can be used: i=m Pmotor br i + Pmotor br i +1 i =1 2 Wbr motor ( t i +1 - t i ) The following is obtained by sub-dividing into 10 segments: Wbr -1739 Ws Siemens AG SIMOVERT MASTERDRIVES - Application Manual 411 4 Information for specific applications 09.99 i t Pmotor motor Pmotordt 0 0 -155.93 0 1 2.69 -131.64 -371.91 2 5.38 -110.98 -685.69 3 8.07 -92.94 -949.42 4 10.76 -76.85 -1169.02 5 13.44 -62.21 -1348.87 6 16.13 -48.63 -1492.21 7 18.82 -35.85 -1601.47 8 21.51 -23.61 -1678.37 9 24.20 -11.72 -1724.06 10 26.89 0.00 -1739.22 Table to numerically calculate the braking energy Diagrams for motor output and motor speed when braking with a constant motor torque Motor output for tbr=26.9 s 0 0 5 10 15 20 25 30 -20 -40 -60 -80 -100 -120 -140 -160 Time in s 412 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 4 Information for specific applications Motor speed for tbr=26.9 s 1600 1400 1200 1000 800 600 400 200 0 0 5 10 15 20 25 30 Time in s Siemens AG SIMOVERT MASTERDRIVES - Application Manual 413 4 Information for specific applications 09.99 Braking with constant deceleration and subsequent torque limiting When braking with constant deceleration, the motor torque increases as the motor speed approaches zero. Under certain circumstances, it is practical to limit the torque, for instance, to the rated motor torque. Naturally the braking time is then extended. In order that the brake resistor is optimally utilized, torque limiting should only be used, if the maximum motor output was reached when braking. The maximum motor output when braking occurs at : n= 3 J 2 n max 226.6 2 1500 3 = = 1099.5 TPM 3 60 M load max t br 180 1260.5 17.52 The value, previously calculated when braking with constant deceleration without torque limiting, is inserted for tbr . The associated motor torque is given by: M motor br = M load max ( = 1260.5 ( n nmax )2 - J 2 n max 60 t br 1099.5 2 226.6 2 1500 ) - = -1354.4 Nm 1500 60 17.52 The motor torque at n=0, without torque limiting, is given by: M motor br = - J 2 n max 226.6 2 1500 =- = -2032 Nm 60 t br 60 17.52 Thus, torque limiting can lie between -1354.4 Nm and -2032 Nm. In this particular case, the motor torque should be limited to the rated torque (-1600 Nm) when braking. M motor t1 t br t - Mn Example for torque limiting when braking (in this case to the rated torque) 414 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 4 Information for specific applications Initially, the speed n1 at t1 must be calculated. 1 n1 = n max M load max ( J 2 n max - Mn ) 60 t br 1 226.6 2 1500 ( - 1600) = 877.8 RPM 1260.5 60 17.52 = 1500 Thus, t1 is given by: t1 = t br (1 - n1 877.8 ) = 17.52 (1 - ) = 7.27 s n max 1500 The following is obtained for the extended" braking time: tbr = t1 + J 2 nmax 60 M load max M n = 7.27 + M load max arctan( Mn n1 ) n max 226.6 2 1500 1260.5 877.8 arctan( ) = 19.28 s 1600 1500 60 1260.5 1600 For t0=0, up to time t1, the braking energy is given by: Wbr 1 2 motor 2 t br M mload max nmax t1 4 J 2 n max t = ( (1 - (1 - ) ) - (1 - (1 - 1 ) 2 )) 60 4 t br 2 60 t br t br 0.962 2 17.52 1260.5 1500 7.27 4 226.6 2 1500 2 7.27 2 = ( (1 - (1 - ) )- (1 - (1 - ) )) 60 4 1552 . 2 60 17.52 1552 . = -1032 kWs To calculate the braking energy in the range t1 to tbr the motor output in this range must first be determined. The following is true: Pmotor br = - M n 2 n 60 with n= n max M load max tan(arctan( M load max Mn 60 M load max M n n1 )- (t - t 1 )) n max J 2 n max Mn Siemens AG SIMOVERT MASTERDRIVES - Application Manual 415 4 Information for specific applications 09.99 In this case, the braking energy can only be numerically determined. The following is obtained using the previous approximation and by sub-dividing into 10 segments: Wbr 2 -815 kWs Thus, the total braking energy is given by: Wbr -1032 - 815 = -1847 kWs i t Pmotor motor Pmotordt 0 7.27 -147.05 0 1 8.47 -130.23 -160.11 2 9.67 -114.14 -301.22 3 10.87 -98.66 -424.10 4 12.07 -83.69 -529.40 5 13.27 -69.14 -617.65 6 14.47 -54.92 -689.28 7 15.67 -40.97 -744.66 8 16.87 -27.21 -784.02 9 18.07 -13.57 -807.57 10 19.27 0.00 -815.41 Table to numerically calculate the braking energy 416 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 4 Information for specific applications Motor output and motor torque diagrams Motor output for tbr=19.28 s 0 0 2 4 6 8 10 12 14 16 18 20 -20 -40 -60 -80 -100 -120 -140 -160 Time in s Motor torque for tbr=19.28 s 0 0 2 4 6 8 10 12 14 16 18 20 -300 -600 -900 -1200 -1500 -1800 Time in s Siemens AG SIMOVERT MASTERDRIVES - Application Manual 417 4 Information for specific applications 09.99 4.2 Connecting higher output motors to a drive converter than is normally permissible 4.2.1 General information Motor and drive converter are optimally utilized when the induction motor is assigned to the drive converter, as specified in the Catalog and Engineering Manual. However, in special cases, it may be necessary to use a higher-rating motor than is generally permissible with the particular drive converter, taking into account a poorer utilization level. Reasons could include: * A large motor is only to be operated under no-load conditions for test purposes * The motor is only operated at partial load, e.g. derating factors up to 0.5 are obtained for Mpermissible/Mn for a speed control range of 1:100 and utilization to temperature rise class B. In cases such as these, a standard motor/drive converter assignment could result in a drive converter with an excessive rating and thus far higher costs. The following must be observed when connecting a higher-rating induction motor to the drive converter than is permissible according to the Catalog or Engineering Manual: * Higher current peaks are obtained due to the lower leakage inductance of higher-rating motors, which under certain circumstances, could cause the drive converter to be tripped by overcurrent. Thus, the maximum drive converter current must be reduced corresponding to the higher rated motor current. Another possibility is to compensate the low leakage inductance using an additional output reactor, or to increase the leakage inductance using / changeover. * At constant flux, the motor power factor deteriorates in partial load operation. The transferrable drive converter active power is then decreased, especially for motors with a high magnetization current component. Thus, in the partial load range, a larger motor cannot generate the same shaft output as the adapted smaller motor. * The motor data, required when vector control is used, may, under certain circumstances, no longer be able to be set when a larger motor is used. Thus, a smaller motor must be parameterized. The motor model values are correctly set via the motor identification function. * If the motor is only to be essentially operated under no-load conditions, it mainly draws reactive current. This reactive current is not supplied from the line supply, but from the DC link capacitors. For reasons of stability, the reactive current component should not be more than approx. 60% of the rated drive converter current. 418 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 4 Information for specific applications Reduction factor if the leakage inductance of the motor is too low As long as the motor leakage reactance does not fall below approximately 10% of the rated drive converter impedance, operation is uncritical. This is the case for the standard induction motors listed in the Catalog and Engineering Manual. When connecting a higher-rating motor without taking any measures (refer below), the following rated motor current limit should not be exceeded in order to prevent higher peak currents: I motor n 1.36 I U n Further, the maximum drive converter current should be reduced as follows: I V max = 136 . IV n - 0.6 2 IV n ( I motor n IV n - 1) for I V n I motor n 136 . IV n Compensating the leakage inductance using an additional output reactor If the motor leakage inductance is to be raised to at least 10% of the drive converter rated impedance using an additional output reactor, then the following must be true: X motor + X reactor = 01 . ZV n = 01 . UV n 3 IV n If the motor leakage is assumed to be 10%, then the reactor inductance is given by: Lreactor 0.1 U V n 10 3 3 314 ( 1 IV n - 1 I motor n ) [mH] As the standard output reactors are designed for approx. 1.2 to 0.6% of the rated drive converter impedance, then, for example when connecting a motor with twice the rating as normally permissible, approximately 4 to 8 standard output reactors would be required. Increasing the leakage inductance using / motor winding changeover The leakage inductance is increased by a factor of 3 using / changeover of the motor winding. Thus, a motor with 3 times the output normally permissible can be fed from the converter under partial load conditions. However, the permissible motor output is then only one third of the rated motor output. Siemens AG SIMOVERT MASTERDRIVES - Application Manual 419 4 Information for specific applications 09.99 Permissible motor torque The following approximation is valid for the permissible motor torque referred to the rated motor torque for a higher-rating motor: M perm. Mn I I n ( I V perm. I mot n )2 - ( 1- ( I I mot n I n I mot n )2 )2 I V perm. : permisible drive converter current at the operating point I mot n : rated motor current I : I n : motor magnetizing current at the operating point rated motor magnetizing current Example 1 A motor with 300% output of that normally permissible in a circuit configuration (660 V) is to be fed from a 380 V drive converter. The rated motor current in the circuit configuration should be 3 IU n and in the circuit configuration I V perm. = I V n IV perm. I mot n = 3 I V n . Under the assumption that: 1 3 and with I I n 3 (field weakening as a result fof the circuit) the permissible motor torque is given by: M perm. Mn 1 3 1 1 I n 2 - ( ) 3 3 I mot n 1- ( I n I mot n )2 = 1 3 The possible output at rated speed is therefore 1/3 of the rated motor output. 420 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 4 Information for specific applications Example 2 A 4-pole 7.5 kW motor is to be fed from a 5.5 kW drive converter as fan/pump drive. The continuous drive converter current (rated drive converter current) is fully utilized. Rated motor current: 15.4 A Rated magnetizing current: 53% Rated converter current: 13.2 A The condition I motor n 136 . I V n is in this case fulfilled. With I = I n the permissible motor torque at rated speed is given by: M perm. Mn ( 13.2 2 ) - 0.532 15.4 = 0.79 1 - 0.53 2 The output at rated speed is then 0.79 7.5 kW = 5.9 kW . 5.5 kW can be achieved as fan/pump drive using the adapted 4-pole 5.5 kW motor connected to the 5.5 kW drive converter. Siemens AG SIMOVERT MASTERDRIVES - Application Manual 421 4 Information for specific applications 4.2.2 09.99 Operating a 600 kW motor under no-load conditions After service work, traction motors are to be checked for symmetrical (balanced) phase currents under no-load conditions and mechanical running characteristics. A motor is accelerated up to maximum speed using a PWM drive converter connected to the 400 V line supply. Motor data Rated output Pn = 600 kW Rated voltage Vn = 1638 V Rated current In = 266 A Rated speed nn = 1670 RPM Rated frequency fn = 85 Hz Max. speed nmax = 3145 RPM Max. frequency fmax = 160 Hz Leakage inductance L = 2.04 mH Rotor resistance R2 = 0.0691 No-load current at the rated operating point I n = 85 A Estimating the required drive converter output according to the no-load current The motor voltage at 50 Hz is: V50 Hz 50 1638 = 963 V 85 As the drive converter can only supply 400 V, the magnetizing current is given by: I 400 400 I n = 85 = 35 A 963 963 The magnetizing current component should not exceed 60% of the rated drive converter current. Thus, the drive converter current is given by: IV n = 100 35 = 58.3 A 60 According to this criterium, then at least a 30 kW drive converter with IU n =59 A is required. 422 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 4 Information for specific applications Estimating the stall torque at 160 Hz The motor stall torque can be approximated as follows: M stall n M n sn s K ( + ) 2 s K sn With sn = n0 - nn 1700 - 1670 = = 0.0176 1700 n0 R2 R2 0.0691 sK = = = = 0.0634 X 1 + X 2 2 f n L 2 85 2.04 10 -3 The following is obtained M stall n 3430 0.0176 0.0634 ( + ) = 6654 Nm 2 0.0634 0.0176 The motor is fed from a drive converter with constant voltage from 50 Hz to 160 Hz. The following is obtained for the stall torque at 160 Hz: M stall (160 Hz ) M stall n ( Vmax 2 f transition 2 400 2 50 2 ) ( ) = 6654 ( ) ( ) = 113 Nm V50 Hz f max 963 160 Thus, with a 1.3 safety margin from the stall torque, only 87 Nm is available for acceleration. This represents 2.54% of the rated motor torque. The associated accelerating power is thus given by: Pb = M b n max 87 3145 = = 28.6 kW 9550 9550 A 37 kW drive converter is to be used (6SE7027-2ED61 with IV n=72 A). Permissible motor leakage inductance The permissible motor leakage inductance is given by: L 0.1 U V n 10 3 3 314 I V n = 0.1 400 10 3 3 314 72 . mH = 102 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 423 4 Information for specific applications 09.99 The motor leakage inductance of 2.04 mH is therefore sufficiently high (this is a function of the high motor rated motor voltage). Closed-loop control type and setting Closed-loop frequency control is preferable due to the improved stability. When parameterizing the system, a 6-pole 30 kW motor (1LA6 223-6) can be used as basis. 424 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 4 Information for specific applications 4.3 Using a transformer at the drive converter output 4.3.1 General information Normally, the rated motor voltage and the maximum drive converter output voltage are harmonized with one another, thus ensuring optimum drive converter utilization. Occasionally, the rated motor voltage and maximum drive converter output voltage are not the same. In this case, it may be necessary to adapt the voltage using an output transformer, e.g.: * Operating processing machines with a lower line supply voltage, e.g. grinding machines, which up until now, were fed from a motor-generator set. * To step-up the motor voltage to keep the cable losses low if long motor feeder cables are used. The following must be observed when using a transformer at the drive converter output: * If there is no sinusoidal filter between the drive converter and transformer, the transformer must be designed for operation at the converter output. Contrary to a transformer connected to the line supply, the frequencies are higher (pulse frequency), and the current spikes. In order to prevent the effects of saturation due to the high current spikes, and as a result of possible current dissymmetries (i.e unbalance), the inductance of the transformer must be utilized to a lower level than otherwise. * DC current braking is no longer possible as the transformer can't conduct DC current. * The achievable motor starting torque is generally lower than if a transformer was not used, as the transformer can only effectively transform above a specific minimum frequency. For rated torque starting, the transformer must be designed to operate, for example, at the rated motor slip frequency. * The drive converter must additionally supply the losses as well as the transformer magnetizing current. * When using an autotransformer, it is not permissible to ground the transformer neutral (star) point, as this would represent a ground fault for the drive converter. An isolating transformer is preferable if long motor feeder cables are used, as otherwise, the influence of the cable capacitance with respect to ground would be significantly reduced (lower charging currents due to the almost sinusoidal voltages, even with respect to ground on the motor side). Siemens AG SIMOVERT MASTERDRIVES - Application Manual 425 4 Information for specific applications * 09.99 When a sinusoidal filter is used, the maximum output voltage is only approx. 85% of the line supply voltage. The transformer ratio must take this into account. Depending on the drive converter output, it may even be necessary to de-rate it due to the high pulse frequency required (6 kHz at 380 to 460 V). 426 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 4 Information for specific applications 4.3.2 Operating a 660 V pump motor through an isolating transformer A 23 kW pump motor, located in a well, is connected to a 30 kW drive converter installed 1 km away in an equipment room. In order to keep the motor cable losses as low as possible, the motor is connected in a circuit configuration at 660 V. In order to adapt the maximum 380 V drive converter output voltage, an isolating transformer is used. In this case, it is not cost-effective to use a 660 V drive converter due to the low motor output. Further, a sinusoidal filter is required between the drive converter and isolating transformer due to the 1000 m feeder cable and as a result of the high 660 V motor voltage (a third-party motor is being used). 1000 m feeder cables are permissible as the isolating transformer de-couples the motor cable ground capacitances. The isolating transformer ratio was selected to compensate the 15% lower output voltage when using the sinusoidal filter. De-rating is not required for the 30 kW drive converter at a 6 kHz pulse frequency. 30 kW drive converter Approx. 1 km cable length 380 V Sinusoidal filter ~ ~ 6SE70260ED61 Isolating transformer 323/660 V, 33 kVA C, D IM 660 V, 23 kW pump motor Block diagram Siemens AG SIMOVERT MASTERDRIVES - Application Manual 427 4 Information for specific applications 4.3.3 09.99 Operating 135V/200 Hz handheld grinding machines through an isolating transformer Handheld grinding machines equipped with three-phase motors are to be fed from a drive converter which is connected to a 380 V supply. In this case, the converter acts as three-phase supply with a fixed voltage and frequency. A maximum of 9 motors can be connected through sockets. The motors can either run individually or all together. An isolating transformer is used to match the voltage. Further, a sinusoidal filter is used as the motors are not designed for converter operation. In addition, EMC disturbances/noise via the motor cables are significantly reduced. Ground currents flowing through cable capacitances are essentially eliminated using the isolating transformer. Motor data Rated voltage Vn = 135 V Rated frequency fn = 200 Hz Rated speed nn = 6000 RPM Rated output/rated current of the motors to be connected: 2 motors with 260 W / 1.41 A 6 motors with 1000 W / 5.6 A 1 motor with 1800 W / 8.6 A It is assumed, that an 1800 W motor will be switched-in, in addition to 2 x 260 W motors and 6x1000 W motors running at rated load. The starting current (inrush current) is 600% In. Thus, the maximum current is given by: I max 2 1.41 A + 6 5.6 A + 6 1 8.6 A = 88 A If all of the motors are operated at rated load, then the following is true: I n total 2 1.41 A + 6 5.6 A + 1 8.6 A = 45 A As the currents are geometrically summed, the algebraic addition made here means that the calculation is on the safe side. The 150% overload capability of the drive converter can be utilized for the maximum current when the highest-rating motor starts. 428 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 4 Information for specific applications Thus, the basic drive converter load current must be at least IU G I max 88 = = 24.5 A . u 15 . 323 / 135 15 A 15 kW drive converter, 6SE7023-4EC61 with 30.9 A base load current is selected. Thus, there is sufficient reserve for the magnetizing current and the transformer losses. The transformer ratio takes into account that only 85 % of the maximum drive converter output voltage can be reached due to the sinusoidal filter. De-rating is not required for the 15 kW drive converter and 6 kHz pulse frequency. 15 kW drive converter Transformer 323/135 V, 12 kVA 380 V 6SE70234EC61 Sinusoidal filter C, D Motor sockets 2 x 260 W, 6 x 1000 W, 1 x 1800 W Block diagram Siemens AG SIMOVERT MASTERDRIVES - Application Manual 429 4 Information for specific applications 4.4 Emergency off 4.4.1 General information 09.99 According to VDE 0113, stop functions fall under the following categories: * Category 0 Uncontrolled shutdown by immediately disconnecting the power feed (the motor coasts down). * Category 1 Controlled shutdown, whereby power is still fed to the drive so that it can shutdown in a controlled fashion. The power is only disconnected once the drive comes to a standstill. * Category 2 Controlled shutdown, whereby the power is still connected even after the drive comes to a standstill. Further, the following additional requirements are valid for emergency off: * Emergency off has priority over all other functions * The power should be disconnected as quickly as possible * It is not permissible that a reset initiates a restart Emergency off must either be effective as a category 0- or category 1 stop function. The actual category must be defined as a result of the machine risk evaluation. For an emergency off function, stop category 0, only permanently-wired (hardwired) electromagnetic components may be used (e.g. neither the drive converter electronics nor PLC controls can be used). For an emergency off function, stop category 1, the power supply must be reliably disconnected, and as for category 0, by just electromechanical components. 430 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 4 Information for specific applications Emergency off via a category 1 stop function (fast stop) In this case, shutdown is initiated with an emergency off command using the off 3" command (fast stop with a suitably selected ramp function). The line contactor is opened to disconnect the power after the drive comes to a standstill. In addition, a time relay is provided to positively disconnect the power so that the line contactor is reliably opened, even if the drive converter electronics develops a fault. When the delay expires, the external fault" command is also output, in order to prevent restarting without first acknowledging the fault. Fast shutdown is only possible if the drive is equipped with a pulsed resistor or regenerative feedback. Fast shutdown does not function if the drive converter or line supply develops a fault. Thus, normally safety-critical drives (e.g. cranes) always have a mechanical brake, which can brake the drive to standstill within a specific time. Further, using this brake, the drive is held at standstill when the electronics fails even if the DC link voltage is still available (the DC link voltage only decays slowly). Emergency off via a category 0 stop function (electrical shutdown) In this case, the system must be shutdown (into a no-voltage condition) as quickly as possible (the motor coasts down or is mechanically braked). There are two methods of achieving this: a) Disconnecting the line supply voltage in front of the drive converter For safety reasons, it is not permissible to open the line contactor using the internal drive converter electronics (the drive converter electronics could fail). Thus, the line contactor is directly opened via the emergency off button/switch. In order to prevent switching under load, an external fault" command is simultaneously output to inhibit the drive pulses (processing time: 4T0). This also prevents the drive starting without first acknowledging the fault. b) Disconnecting the line supply voltage in front of the drive converter and disconnecting the motor If only the line supply voltage is disconnected in front of the converter (feeding the drive converter), the DC link still remains charged for a specific time. Thus, voltage is still present at the motor terminals. In order to ensure that the voltage at the motor terminals is zero, a motor contactor located between the motor and drive converter must open. The motor contactor must be closed before the drive converter is powered-up, and it may only be opened if there is no current flowing through the motor. This can be realized using the main contactor checkback signal. Siemens AG SIMOVERT MASTERDRIVES - Application Manual 431 4 Information for specific applications 09.99 Contactor opening when the line supply fails When the line supply fails, the contactors can, under certain circumstances, drop-out (open), before the drive converter monitoring function can output a pulse inhibit command. This is the case, for example, in the partial load range, during short line supply dips, or when kinetic buffering becomes active. Thus, in this case, the contactor opens under load. Especially for motor contactors, switching under load must be prevented. If the contactor switches under load at low output frequencies, a quasi DC current could be interrupted which could result in arcing. To prevent this happening, the contactors (control circuit) must be supplied from an uninterruptable power supply. 432 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 4 Information for specific applications Auxiliary voltage e.g. 230 V AC 24 V DC external auxiliary voltage -K4 -K1 -K4 -x1: U1/ L1 V1/ L2 W1/ L3 -x9: 1 2 -x101: 16 -x9: 4 On/off -x101: 13 -x9: 5 -x101: 17 6SE70 K2 (off 3) -x101: 13 -x101: 18 K3 (ext. fault) -x101: 13 -x2: U2/ T1 V2/ T2 W2/ T3 M 3~ Emergency off K2 K2 K3 K3 K4 Delayed drop-out, time somewhat longer than the fast stop ramp Example of an emergency off circuit using a category 1 stop (fast stop and then the power is disconnected) Siemens AG SIMOVERT MASTERDRIVES - Application Manual 433 4 Information for specific applications Auxiliary voltage e.g. 230 V AC 09.99 24 V DC ext. auxiliary power supply -K3 -K1 -K3 -x1: U1/ L1 V1/ L2 W1/ L3 -x9: 1 2 -x101: 16 -x9: 4 ON/off -x101: 13 -x9: 5 -x101: 17 6SE70 -x2: U2/ T1 V2/ T2 K2 (ext. fault) -x101: 13 W2/ T3 M 3~ Emergency off K2 K2 K3 Example of an emergency off circuit using a category 0 stop (the line supply voltage is disconnected) 434 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 4 Information for specific applications Auxiliary voltage e.g. 230 V AC 24 V DC ext. auxiliary power supply -K4 -K1 -K4 -x1: U1/ L1 V1/ L2 W1/ L3 -x9: 1 2 -x9: 4 -x101: 16 -x9: 5 -x101: 13 -x101: 17 6SE70 On/off K3 (ext. fault) -x101: 13 -x101: 18 K2 checkback signal -x101: 13 -x2: U2/ T1 V2/ T2 W2/ T3 -K2 M 3~ Emergency off K3 K3 K4 An example of an emergency off circuit using a category 0 stop. (the line supply voltage is disconnected and the motor is disconnected). Caution: Observe the maximum switching capability of contact X9: 4.5. Siemens AG SIMOVERT MASTERDRIVES - Application Manual 435 4 Information for specific applications 09.99 4.5 Accelerating- and decelerating time with a constant load torque in the field-weakening range 4.5.1 General information The prerequisite for the calculation is a constant motor torque in the constant-flux range and a constant motor output in the field-weakening range (e.g. accelerating along the current limit). Thus, the motor torque is given by: M motor = M motor max = const M motor = M motor max (constant flux range) nn n (field-weakening range, n nn) Mmotor Pmotor Accelerating torque M load nn n max n Example for starting in the field-weakening range Under these assumptions, the accelerating time (ta) and the decelerating time (tbr) can be calculated as follows. 436 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 4 Information for specific applications * Case 1 ( M load = 0 , i.e. pure high-inertia drive (flywheel)) In the constant-flux range, with 0 n nn the following is true: t a , br * J total 2 nn = 60 M motor max In the field-weakening range, with nn n nmax the following is true: t a , br 2 * (n max - nn2 ) J total = 60 M motor max nn * Case 2 ( M load 0 , e.g. friction present) In the constant-flux range, with 0 n nn the following is true: t a , br = * J total 2 nn * 60 ( M motor max m M load ) In the field-weakening range with nn n nmax the following is true: t a , br * M motor max nn J total 2 n -n = ( n * max + ln *2 60 M load M load * M motor max m M load ) n max * M motor max m M load nn The lower sign in the last two formulas is true when the drive brakes * M load Constant load torque referred to the motor [Nm] * J total Total moment of inertia referred to the motor [kgm2] M motor max Motor torque in the constant-flux range [Nm] nn Rated motor speed (start of field weakening) [RPM] nmax Max. motor speed in the field-weakening range [RPM] Siemens AG SIMOVERT MASTERDRIVES - Application Manual 437 4 Information for specific applications 4.5.2 09.99 Braking time for a grinding wheel drive The motor of a grinding wheel drive is operated in the field-weakening range from 50 Hz to 60 Hz. The minimum braking time is to be determined when a drive converter with brake resistor is used. The load torque when braking is assumed to be zero (friction has been neglected). There is no gearbox. Drive data Rated motor output Pmotor n = 11 kW Rated motor current Imotor n Rated motor torque Mmotor n = 36 Nm Rated motor speed (2 pole) nn = 2915 RPM Motor efficiency motor =0.87 Motor moment of inertia Jmotor = 0.034 kgm2 Ratio of the stall torque to the rated motor torque Mstall/Mn = 2.6 Maximum speed of the grinding wheel nmax = 3600 RPM Moment of inertia of the grinding wheel Jwheel = 0.34 kgm2 Rated drive converter output PV n = 11 kW Max. drive converter current IU max = 34.8 A External brake resistor rating P20 = 20 kW Peak braking power 1.5 P20 = 30 kW = 21.4 A Determining the maximum motor torque The maximum possible motor torque at the maximum drive converter current is: M motor max M motor n I V2 max - I 2 n 2 2 I motor n - I n With I n = 0.41 I motor n the following is obtained: M motor max 36 438 34.8 2 - 0.412 21.4 2 = 62 Nm 21.4 2 - 0.412 214 . 2 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 4 Information for specific applications In the field-weakening range at nmax and at constant motor output, the motor torque is then only given by: M motor ( 3600) = M motor max nn 2915 = 62 = 50.2 Nm 3600 n max On the other hand, the permissible motor torque at nmax and 130% margin for the stall torque is given by: M motor perm. ( 3600) = 2.6 M motor n 1.3 ( nn 2 2.6 36 2915 2 ) = ( ) = 47.2 Nm n max 1.3 3600 Thus, the maximum permissible motor torque is reduced to: M motor max = M motor perm. ( 3600) n max 3600 = 47.2 = 58.3 Nm 2915 nn Thus, the drive converter is not fully utilized when braking. Determining the braking times Total moment of inertia: J total = J motor + J wheel = 0.034 + 0.34 = 0.374 kgm2 Braking time in the field-weakening range: t br 1 2 J total (n max - nn2 ) 0.374 (3600 2 - 29152 ) = = = 0.51 s 60 M motor max nn 60 58.3 2915 Braking time in the constant-flux range: t br 2 = J total 2 n n 0.374 2 2915 = = 1.96 s 60 M motor max 60 58.3 Thus, the total braking time is given by: t br total = t br 1 + t br 2 = 0.51 + 1.96 = 2.47 s Siemens AG SIMOVERT MASTERDRIVES - Application Manual 439 4 Information for specific applications 09.99 Determining the maximum braking power and braking energy The maximum braking power is given by: Pbr max = M motor max nn 9550 motor = 58.3 2915 0.87 = 15.48 kW 9550 The brake resistor with P20=20 kW and a peak braking power of 30 kW is more than adequately dimensioned. The braking energy corresponds to the area in the braking diagram. Braking diagram P br P br max tbr 1 Wbr = Pbr max t br 1 + tbr 2 Pbr max t br 2 2 t = 15.48 0.51 + 15.48 1,96 = 23.07 kWs 2 With P Wbr Pbr cont . = 20 T 4.5 440 (with an external brake resistor) Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 4 Information for specific applications The permissible period is given by: T Wbr 4.5 23.07 4.5 = = 5.2 s P20 20 Thus, braking can be realized every 5.2 s. Siemens AG SIMOVERT MASTERDRIVES - Application Manual 441 4 Information for specific applications 4.6 09.99 Influence of a rounding-off function Acceleration steps are eliminated, and therefore softer positioning achieved, using a rounding-off function for a traversing characteristic. We will now investigate the influence of such a rounding-off function on the motor- and drive converter dimensioning. These investigations are based on the square-law rounding-off, the sin2 rounding-off, as well as rounding-off for jerk-free motion. These various rounding-off functions are compared with the behavior without rounding-off, i.e. acceleration as a step function. Square-law rounding-off function For a square-law rounding-off function, acceleration is entered as a trapezoid. This results in a square-law characteristic of the velocity during linearly increasing or linearly decreasing acceleration. a a max with 0 k 1 k tb k 2 tb t 2 v vmax 1 2 3 t tb 442 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 4 Information for specific applications Acceleration in the areas 1, 2, 3: a1 = 2 a max t k tb a 2 = a max a 3 = 2 a max tb - t k tb The following is true for the ratio between maximum acceleration and maximum velocity: a max = v max k t b (1 - ) 2 Velocity in areas1, 2, 3: v1 = a max 2 t k tb square-law v2 = a max k t b k tb ) + a max ( t - 4 2 linear a k v3 = a max t b (1 - ) - max ( t b - t ) 2 2 k tb square-law For k=0 (0% rounding-off), areas 1 and 2 are omitted. Acceleration is a step function. For k=1 (100% rounding-off), area 2 is omitted. Acceleration is triangular. The relationships without rounding-off (step-function acceleration) will now be compared with 100% rounding-off (triangular acceleration). This assumes, traversing motion with the same travel, same travel time and the same maximum velocity. Siemens AG SIMOVERT MASTERDRIVES - Application Manual 443 4 Information for specific applications 09.99 a a k=1 max k=0 t v v max k=1 k=0 t tb tv t tot. Comparison of the traversing curve with and without rounding-off The area under the traversing curve v corresponds to the travel stotal. In this case, tb=tv. With a max = v max k t b (1 - ) 2 the following is obtained if k=1: a max = 2 444 v max tb Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 4 Information for specific applications With the same travel stotal, the same travel time ttotal and the same maximum velocity vmax, the maximum acceleration for 100% rounding-off is exactly double. For a pure accelerating drive, the motor torque would also then double (the motor must be appropriately dimensioned) and, corresponding to this torque, also the motor current (drive converter dimension). However, the maximum motor torque only occurs at half the maximum motor speed. In the following, for the previously described traversing motion, the motor torque characteristics are shown as a function of the speed using a pure accelerating drive. The motor torque must lie below the permissible limits (in this case, the limiting curve for an 1FK6 motor). The motor torque for 100% rounding-off (k=1) is compared with the motor torque without rounding-off (k=0). 40 M per. 35 30 k=1 M in Nm 25 20 15 k=0 10 5 0 0 500 1000 1500 2000 2500 3000 Motor speed in RPM Comparison of the motor torque with and without rounding-off Siemens AG SIMOVERT MASTERDRIVES - Application Manual 445 4 Information for specific applications 09.99 Sin2 rounding-off For a sin2 rounding-off, the velocity is entered according to a sin2 function. For the acceleration, a sinusoidal function is obtained with twice the frequency. a a max t tb v v max t Velocity: v = v max sin 2 ( t ) for 0 t tb with = 446 2 tb Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 4 Information for specific applications Acceleration: a = a max sin( 2 t ) for 0 t tb with a max = v max 2 The relationships without rounding-off (step-function acceleration) will now be compared with sin rounding-off (sinusoidal acceleration). It is assumed that the traversing operation has the same travel, the same travel time and the same maximum velocity. a a 2 sin rounding-off max without rounding-off t v sin2 rounding-off v max without rounding-off t tv tb t total Comparison of the traversing curve, with and without rounding-off Siemens AG SIMOVERT MASTERDRIVES - Application Manual 447 4 Information for specific applications 09.99 The area under the traversing curve v corresponds to the travel stotal. In this case, tb=tv. With a max = v max = v max 2 tb a factor of /2 is obtained with respect to the acceleration value without rounding-off. For the same travel stotal, the same travel time ttotal and the same maximum velocity vmax, the maximum acceleration for sin2 rounding-off is therefore greater by a factor /2. For a pure accelerating drive, the motor torque would thus be increased by this factor (the motor must be appropriately dimensioned), and, corresponding to this torque, also the motor current (the drive converter must be appropriately dimensioned). However, the maximum motor torque only occurs at half the maximum motor speed. The previously described traversing motion is illustrated in the following using a pure accelerating drive, showing the motor torque as a function of the speed. The motor torque must lie within the permissible limits (in this case, the limiting curve for an 1FK6 motor). The motor torque for sin2 rounding-off is compared with the motor torque without rounding-off. 40 Mper. 35 M in Nm 30 with sin2 -rounding-off 25 20 15 without rounding-off 10 5 0 0 500 1000 1500 2000 Motor speed in RPM 2500 3000 Comparison with and without rounding-off 448 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 4 Information for specific applications Rounding-off for jerk-free motion For jerk-free motion, the acceleration must be specified so that the derivative at the transitions is 0. For example, this is achieved with a sin 2 function for the acceleration. a amax tb t v v max t Acceleration: a = amax sin 2 ( t ) for 0 t tb with = tb Siemens AG SIMOVERT MASTERDRIVES - Application Manual 449 4 Information for specific applications 09.99 Velocity: v = v max ( t 1 - sin( 2 t )) tb 2 tb for 0 t tb with v max = 1 a t 2 max b The relationships without rounding-off (step-function acceleration) with rounding-off for jerk-free movement (sin2-function acceleration) will now be compared. It is assumed that the traversing operation has the same travel, the same travel time and the same maximum velocity. a amax jerk-free rounding-off without rounding-off t v v max jerk-free rounding-off without rounding-off t tb tv t total Comparison of the traversing curve with and without rounding-off 450 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 4 Information for specific applications The area under the traversing curve v corresponds to the travel stotal. In this case, tb=tv. With a max = 2 v max tb the same travel stotal, the same travel time ttotal and the same maximum velocity vmax results in twice the acceleration value with respect to the traversing curve without rounding-off. For a pure accelerating drive, the motor torque would thus be increased by this factor (the motor must be appropriately dimensioned), and, corresponding to this torque, also the motor current (the drive converter must be appropriately dimensioned). However, the maximum motor torque only occurs at half the maximum motor speed. The previously described traversing motion is illustrated in the following using a pure accelerating drive, as an example for the motor torque characteristic as a function of the speed. The motor torque must lie within the permissible limits (in this case, the limiting curve for an 1FK6 motor). The motor torque for sin2 rounding-off is compared with the motor torque without rounding-off. 40 M per. 35 30 jerk-free rounding-off M in Nm 25 20 15 without rounding-off 10 5 0 0 500 1000 1500 2000 2500 3000 Motor speed in RPM Comparison of the motor torque, with and without rounding-off Siemens AG SIMOVERT MASTERDRIVES - Application Manual 451 4 Information for specific applications 09.99 Summary If, for a positioning drive, the same traversing sequence is investigated with and without roundingoff, then, for the same positioning time, and the same maximum velocity, the traversing curve with rounding-off, requires a higher acceleration and deceleration. Thus, the motor torque and therefore the motor current increases. Due to the higher motor torque, the RMS torque also increases. Under certain circumstances, a larger motor and a larger drive converter must be used. By appropriately increasing the acceleration- and deceleration times, acceleration and deceleration can be reduced. Naturally, the positioning time then increases. For square-law rounding-off and sin2 rounding-off, the acceleration steps are avoided, but completely jerk-free motion is not achieved. This is only achieved when the sin2-type acceleration function is applied. This type of acceleration setpoint doubles the value for the acceleration, and also results in the highest increase in the RMS value. 452 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 4 Information for specific applications 4.7 Possible braking torques due to system losses 4.7.1 General information Normally, when braking, the braking energy is either dissipated in a brake resistor, fed back into the line supply, or, when using DC current braking, converted in the motor itself. There are only a very limited number of ways of storing the braking energy in the DC link. Below is an example of a 4-pole 55 kW motor connected to a 55 kW drive converter with a 400 V supply voltage. With J Mot = 0.79 kgm 2 motor moment of inertia CDClink = 9 mF drive converter DC link capacitance the kinetic energy W kin, to shutdown the motor and the energy W C which can be stored in the DC link capacitor is given by: Wkin Mot = WC = 2 n Mot 2 1 1 2 1475 2 J Mot ( ) = 0.79 ( ) = 9424 Ws 2 60 2 60 1 1 2 2 C DClink (U DClink 9 10 -3 (800 2 - 530 2 ) = 1616 Ws max - U DClink n ) = 2 2 Using the energy stored in the DC link capacitor, not even the motor alone can be shutdown without taking into account the losses. Thus, in the following, the possible braking torque is determined, taking into account the loss equation. The following diagram makes the situation clear. P V Inverter Line supply P V react. Inverter Rectifier P M P V const. Siemens AG SIMOVERT MASTERDRIVES - Application Manual P br V motor 453 4 Information for specific applications 09.99 The motor braking power may not exceed the sum of the losses in the motor, in the output reactor, in the inverter and in the power supply (constant losses). Thus, the following must be true: PV Mot + PV react + PV inv . + PV const Pbr With Pbr = M br n Mot 9550 the following is obtained: M br ( PV Mot + PV react .r + PV inv . + PV const ) 9550 n Mot There is somewhat of a problem in so much that the inverter losses are load-dependent, and the motor- and reactor losses, load- and speed-dependent. As the speed decreases, the iron losses of the motor and reactor decrease while the current-dependent losses of the motor, reactor and inverter increase due to the increasing braking torque. As a first approximation, it is assumed that the sum of the losses is constant. As the expected braking torques at rated speed are low, an operating point close to the no-load motor is assumed. In this case, the magnetizing current can be considered to be the motor current. Thus, useable results can be obtained at rated speed using the formula above for the braking torque. The calculation becomes increasingly more inaccurate as the speed decreases. For I Mot I mag , the losses at rated speed can be estimated as follows: PV Mot 0.4...0.5 PV Mot n = 0.4...0.5 PMot n ( PV Dr PV Cu n ( I mag I Dr n 1 Mot n - 1) ) 2 + PV Fe n PV inv . ( PV D n + PV Sch n 0.93 500 2 I mag ( ) ) 3 400 IU n The constant losses comprise of the losses of the power supply, switching losses in the DC link capacitors, and for small units, additionally the losses due to the DC fan. 454 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 4 Information for specific applications Examples for the losses and braking torques at rated speed For a 4-pole 5.5 kW motor with output reactor connected to a 5.5 kW drive converter with 400 V supply voltage, the losses are given by: P V 0.58 kW or for the rated torque at rated speed: M br P V 9550 nn = 0.58 9550 . Nm = 382 1450 This corresponds to 10.6% of the rated torque. For a 4-pole 55 kW motor with output reactor connected to a 55 kW drive converter with 400 V supply voltage, the losses are given by: P V 2.07 kW and, at rated speed the torque is: M br P V 9550 nn = 2.07 9550 = 13.4 Nm 1475 This corresponds to 3.8% of the rated torque. This means that the possible braking torque decreases with increasing power. Siemens AG SIMOVERT MASTERDRIVES - Application Manual 455 4 Information for specific applications 09.99 Calculating the braking time for a pure high-inertia drive (flywheel drive) In this case M br = - J tot . 2 dn 60 dt For a constant braking torque, a linear braking ramp is obtained with t br = J tot . nmax 30 M br If the equation, derived from the losses, is now used for Mbr M br = P V 9550 n = P 9550 nmax n = M br nmax max nmax n n V then t, as a function of n is given by: J tot . J tot . 2 t=- n dn = (nmax - n2 ) 30 M br nmax nmax nmax 60 M br nmax n max n or t br = J tot nmax 60 M br nmax The motor speed, as a function of time is: 2 n = nmax - 60 M br nmax n max J tot . t The braking time, with the braking torque assumed as 1/n, is precisely half as long as for a constant braking torque. It is assumed, that the Ud max control automatically ensures a down ramp with a square-root characteristic. 456 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 4 Information for specific applications n nmax M br = const. Mbr ~1/n t Example of a braking characteristic for a pure high-inertia drive (flywheel drive) Calculating the braking time for a fan drive In this case M br + M load = - J tot . 2 dn 60 dt The load torque of the fan is given by: M load = M load max ( n n max )2 With M br = M br nmax n max n t, as a function of n, is given by: J tot . 30 nmax n t=- dn M br nmax n max n 2 ) + M load max ( n nmax Siemens AG SIMOVERT MASTERDRIVES - Application Manual 457 4 Information for specific applications 2 J tot . nmax = 30 M load max + 09.99 2 1 a 2 - a nmax + nmax a 2 - a n + n2 (ln - ln ) (a + nmax ) 2 ( a + n) 2 6 a 2 n max - a 1 2 n - a (arctan( ) - arctan( )) a 3 a 3 a 3 or 2 J tot . nmax t br = 30 M load max 2 1 a 2 - a nmax + nmax 2 nmax - a 1 ln + (arctan( ) + ) (a + nmax ) 2 6 a 3 a 3 6 a With a = nmax 3 M br nmax M load max Without taking into account the motor braking torque, the speed, as a function of time, is given by: n= 1 nmax + 1 30 M load max t 2 J tot . n max Zero speed is in this case, only reached for t . 458 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 4 Information for specific applications n nmax without motor braking torque (only M load ) with motor braking torque t Example of a braking characteristic for a fan drive Summary When braking using the losses, only low braking torques can be achieved at the rated operating point (approximately 2 to 10% Mn depending on the motor size). This technique can only be used, where, on one hand short braking times are not important, and where on the other hand, due to large moments of inertia and low friction losses, run-down takes minutes. This occurs, e.g. for large fan drives. In this case, at high speed, the load torque initially has a high braking effect, but which decreases according to an n2 function. Siemens AG SIMOVERT MASTERDRIVES - Application Manual 459 4 Information for specific applications 4.7.2 09.99 Braking time for a fan drive The braking time is to be estimated for a 200 kW fan drive with 4-pole 1LA6 317-4.. motor and 200 kW drive converter connected to 400 V. There is no brake resistor. Drive data Rated motor current Imotor n = 345 A Rated motor torque Mmot n = 1280 Nm Motor efficiency for drive converter operation motor = 0.95 Motor magnetizing current Imag = 107 A Motor moment of inertia Jmot = 4.2 kgm2 Rated drive converter current IU n = 370 A Power requirement at nmax Pmax = 196 kW Maximum fan speed nmax = 1485 RPM Moment of inertia of the fan Jload = 75 kgm2 Determining the losses at nmax with Imot=Imag (values for the drive converter from the Engineering Manual for Drive Converters) PV Mot 0.4 PMot n ( PV react . PV Cu n ( 1 Mot n I mag I react . n PV inv . ( PV D n + PV Sch n PV const 015 . kW - 1) = 0.4 200 ( 1 - 1) = 4.2 kW 0.95 ) 2 + PV Fe n = 0138 . ( 107 2 ) + 0.088 = 01 . kW 370 0.93 500 2 I mag 0.93 500 2 107 ( ) ) = (3.02 + 0.89 ( ) ) = 1 kW 3 400 3 400 370 IU n (fan losses are subtracted) Thus, the sum of the losses is P V = 5.45 kW The braking time is now calculated with these losses which are assumed constant. It is assumed, that the closed-loop Ud max control automatically defines the downramp. 460 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 4 Information for specific applications Maximum load torque: M load max = Pmax 9550 196 9550 = = 1260 Nm nmax 1485 Motor braking torque at nmax: M br nmax = P V 9550 nmax = 5.45 9550 = 35 Nm 1485 Thus, the ratio of the braking torque to the rated motor torque at nmax is: M br nmax M Mot n = 35 = 0.0273 = 2.73% 1280 At nmax, this braking torque is small with respect to the braking effect of the fan load torque. The braking effect due to losses only becomes effective at a lower speed. Calculating the braking time a = nmax 3 M br nmax M load max = 1485 3 35 = 449.74 RPM 1260 2 2 1 a 2 - a nmax + nmax J tot . nmax 2 nmax - a 1 t br = ln + (arctan( ) + ) 30 M load max 6 a (a + nmax ) 2 6 a a 3 3 79.2 14852 = 30 1260 449.74 2 - 449.74 1485 + 14852 1 ln (449.74 + 1485) 2 6 449.74 + 1 2 1485 - 449.74 (arctan( ) + ) = 29.32 s 6 449.74 3 449.74 3 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 461 4 Information for specific applications 09.99 1600 1400 Motor speed in RPM 1200 1000 800 600 400 200 0 0 5 10 15 20 Time in s 25 30 35 40 Braking characteristic for the fan drive Estimating the influence of the energy which can be stored in the DC link capacitor The closed-loop Udmax control limits the DC link voltage to the following value in generator operation: U DClink max = 119 . 2 U line = 119 . 2 400 = 675 V The following is obtained for the energy which can be stored in the DC link capacitor with CDClink=18.8 mF: WC = 1 1 2 2 C DClink (U DClink 18.8 10 -3 (6752 - 5302 ) = 164 . kWs max - U DClink n ) = 2 2 The average power which can be absorbed in the capacitor, during braking is given by: PCm = WC 164 . = = 0.056 kW t br 29.32 This component is low with respect to the total losses of 5.45 kW, and can therefore be neglected in the calculation. 462 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 4 Information for specific applications Comparison with DC braking As a comparison, the possible braking torques using DC current braking will be estimated for this fan drive. The DC current braking torque can be given by: M br M a ( I1 2 ) Ia 1 R '2 s + 22 Xh s with s= n ns slip when braking ns synchronous speed Ia starting current (inrush current) Ma starting torque I1 equivalent three-phase current to generate the braking torque I1 can be converted as follows into a DC braking current Ig: I1 = 2 I 3 g M br ns n Example of a braking torque curve for DC current braking Siemens AG SIMOVERT MASTERDRIVES - Application Manual 463 4 Information for specific applications 09.99 The braking torque, for speeds which are not too low, can be approximated as: M br M a ( I1 2 ns n 2 Ig 1 ) = M a ( )2 s ~ Ia n 3 Ia n n Thus, also when braking using the losses, a braking torque is obtained which is proportional to 1/n. This allows the braking torque to be calculated for the 200 kW motor at rated speed. With I a = 7 I n = 7 345 = 2415 A M a = 2.7 M n = 2.7 1280 = 3456 Nm I g = IU n = 370 A (braking with the rated drive converter current) the following is obtained at rated speed ( nn ns ) 2 370 2 M br 3456 ( ) = 54 Nm 3 2415 The braking torque which can be achieved with the rated drive converter current and DC current braking is therefore somewhat greater than the braking torque calculated using the losses (35 Nm). Estimating the braking time a = nmax 3 M br nmax M load max = 1485 3 54 = 519.68 RPM 1260 2 2 1 a 2 - a nmax + nmax 2 nmax - a J tot . nmax 1 t br ln + (arctan( ) + ) 30 M load max 6 a (a + nmax ) 2 6 a a 3 3 79.2 14852 = 30 1260 519.68 2 - 519.68 1485 + 14852 1 ln (519.68 + 1485) 2 6 519.68 + 1 2 1485 - 519,68 (arctan( ) + ) = 24.1 s 6 519.68 3 519.68 3 A somewhat shorter braking time is obtained due to the higher braking torque. In comparison to braking using the losses, for DC current braking, the total motor braking power is converted in the rotor. In addition, stator copper losses occur due to the DC current supply. 464 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 4 Information for specific applications 4.8 Minimum acceleration time for fan drives with a linear acceleration characteristic 4.8.1 General information The linear acceleration of a fan drive with square-law load torque characteristic is to be investigated. At starting, a breakaway torque, decreasing as square-law, must be additionally taken into account. The following is obtained for the load torque as a function of speed n: M load = M load max ( M load = M load max ( n n max n nmax ) 2 + M LB (1 - n 2 ) n LB )2 0 n nLB n LB n n max M load Mload max M LB Breakaway component n LB n max n Example of a load torque characteristic with breakaway torque The maximum motor torque during acceleration is obtained with MLB<Mload max M Mot max = ( J Mot + J load ) + M load max with = 2 n max 60 t H Thus, the minimum acceleration time is given by: tH = 2 nmax ( J Mot + J load ) 60 ( M Mot max - M load max ) Siemens AG SIMOVERT MASTERDRIVES - Application Manual 465 4 Information for specific applications 09.99 For the given drive converter, the maximum motor torque is obtained from the maximum permissible drive converter current IU max=Imot max: M Mot max = kn = 1 kn = n max nn M Mot n kn IU2 max - ( 2 I Mot n I mag kn 2 - I mag )2 for n max n n constant flux range for n max > n n field-weakening range The maximum motor torque should not exceed 2 Mn for reasons of stability. For operation in the field-weakening range at nmax, sufficient clearance must still be kept to the stall limit: M Mot max M stall n ( n )2 nmax 13 . Thus, the maximum motor torque must lie below the following characteristic: 2 Mn M stall 1.3 n limit ( nn 2 ) n n Where: nlim it = nn 466 M stall 2 M Mot n 13 . Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 4 Information for specific applications We still have to check whether the motor RMS current (corresponds to the drive converter RMS current) is less than or equal to the rated drive converter current when accelerating. The motor RMS current is calculated in a 300 s interval. The following is valid for tH 300 s: 300 s I rms = I 2 Mot tH dt = 0 300 s I Mot k : I 2 Mot 2 dt + I Mot k ( 300 s - t H ) 0 300 s constant motor current at the operating point Mmot=Mload max Imot I mot max Imot k t tH 300 s Example of the motor current characteristic (tH 300 s) If the acceleration time tH is to be greater than 300 s, the interval from tH -300s to tH is used in the calculation for the maximum RMS value during acceleration. The RMS value is then given by: tH I rms max = I 2 Mot t H - 300 s dt 300 s Siemens AG SIMOVERT MASTERDRIVES - Application Manual 467 4 Information for specific applications 09.99 I mot I mot max I mot k 300 s t H t Example for the motor current characteristic (tH > 300 s) In order to calculate the RMS value, the motor current must be determined as a function of the time. The motor current is obtained as follows from the motor torque: I Mot = I Mot n ( I mag 2 I mag 2 1 M Mot 2 ) (1 - ( ) ) kn 2 + ( ) 2 M Mot n I Mot n I Mot n kn kn = 1 n kn = nn for n nn for n > nn The motor torque during acceleration is given by: M Mot = ( J Mot + J load ) + M load (n) After acceleration, the following is true M Mot = M load max For Mload (n) and , the equations shown at the beginning of this text are valid. The following is now inserted for n: n = n max 468 t tH (linear acceleration) Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 4 Information for specific applications The motor current is then known as a function of time. The RMS value is calculated for tH 300 s in 4 ranges: 0 t n LB t n max H breakaway, constant flux range n LB n tH < t n tH n max n max constant flux range nn t < t tH n max H field-weakening range (if available) t H < t 300 s steady-state with Mmot=Mload max For tH >300 s, the last range is eliminated, and the starting point for the calculation is within one of the first 3 ranges. The integral to calculate the RMS value is numerically evaluated using an Excel table. Further, the table of the curve output is used as Excel graphics. i ti ni Mload i Mmot i kni Imot i 2 I Mot dt 0 0 0 MLB .. 1 .. 0 1 .. .. .. t(nLB) nLB t(nn) nn .. .. 2 .. 1 ni /nn tH nmax Mload max Mmot max Imot max m-1 tH nmax Mload max Mload max Imot k m 300 s nmax Mload max Mload max Imot k 2 I rms 300 s Calculation table for tH 300 s The following is formed in the last column: i= m I Mot i-1 + I Mot i i =1 2 ( ) 2 ( t i - t i -1 ) Siemens AG SIMOVERT MASTERDRIVES - Application Manual 469 4 Information for specific applications 09.99 This approximately corresponds to the expression: 300 s I 0 tH 2 Mot dt or I 2 Mot 2 dt + I Mot k (300 s - t H ) 0 A zero is in the first line (i=0). 470 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 4 Information for specific applications 4.8.2 Example The minimum acceleration time for linear acceleration for a fan drive with breakaway torque, decreasing according to a square-law characteristic, is to be calculated. Further, the RMS value of the drive converter current when accelerating, should be checked. The output of the 4-pole fan motor is 90 kW for a 400 V volt supply voltage. A 90 kW drive converter is used. Drive data Rated motor output Pmot n = 90 kW Rated motor torque Mmot n = 581 Nm Rated motor current Imot n = 160 A Magnetizing current Imag = 52.8 A Rated motor speed nn = 1480 RPM Max. speed nmax = 1600 RPM Motor moment of inertia Jmot = 1.6 kgm2 Stall torque Mstall = 1568.7 Nm Load moment of inertia Jload = 200 kgm2 Max. load torque Mload max = 500 Nm Fan breakaway torque MLB = 100 Nm Fan breakaway speed nLB = 200 RPM Rated drive converter current IU n = 186 A Max. drive converter current IU max = 253 A The max. motor torque is given by kn = nmax 1600 = = 1081 . nn 1480 for M Mot max = M Mot n kn IU2 max - ( I mag kn 2 2 - I Mot I n mag )2 581 = . 1081 52.8 2 ) . 1081 = 883.3 Nm < 2 M Mot n 160 2 - 52.8 2 2532 - ( Thus, the minimum acceleration time is given by: tH = 2 nmax ( J Mot + J load ) 2 1600 (16 . + 200) = = 8813 . s < 300 s 60 ( M Mot max - M load max ) 60 (883.3 - 500) Siemens AG SIMOVERT MASTERDRIVES - Application Manual 471 4 Information for specific applications 09.99 Checking the stall limit: M stall nn 2 1568.7 1480 2 ( ( ) = 1032.5 Nm > M Mot max ) = nmax 13 . . 13 1600 In order to calculate the RMS value, the speed range from 0 to nLB is sub-divided into 5 identical sections, the speed range from nLB up to nn, into 10 identical sections, and the speed range from nn to nmax, into 5 identical sections. Time ti is calculated over the speed ni as follows: ti = ni t nmax H Thus, the following calculation table is obtained. i ti 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 ni 0 2,20314175 4,40628349 6,60942524 8,81256698 11,0157087 18,0657623 25,1158159 32,1658695 39,2159231 46,2659766 53,3160302 60,3660838 67,4161374 74,466191 81,5162446 82,8381296 84,1600147 85,4818997 86,8037848 88,1256698 88,1256698 300 M load i 0 40 80 120 160 200 328 456 584 712 840 968 1096 1224 1352 1480 1504 1528 1552 1576 1600 1600 1600 100 64,3125014 37,2500055 18,8125123 9,00002188 7,81253418 21,0125919 40,6126777 66,6127914 99,0129332 137,813103 183,013301 234,613526 292,61378 357,014062 427,814372 441,801933 456,014495 470,452058 485,114622 500,000000 500,000000 500,000000 M mot i 483,298127 447,610629 420,548133 402,11064 392,298149 391,110662 404,310719 423,910805 449,910919 482,311061 521,11123 566,311428 617,911654 675,911908 740,312189 811,112499 825,10006 839,312622 853,750186 868,41275 883,300315 500,000000 500,000000 kn i I mot i 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1,01621622 1,03243243 1,04864865 1,06486486 1,08108108 1,08108108 1,08108108 136,282178 127,779981 121,408307 117,11084 114,83971 114,565661 117,621629 122,196064 128,324764 136,045657 145,394241 156,400536 169,087684 183,471912 199,563359 217,367301 224,078497 230,99668 238,122442 245,456565 253 148,765367 148,765367 2 I Mot dt 0 38405,6206 72606,5335 103941,479 133574,318 162560,407 257579,157 358945,754 469562,303 592747,395 732353,29 892883,165 1079608,38 1298685,75 1557274,69 1863654,47 1928054,86 1996493,28 2069221 2146501,26 2228609,8 2228609,8 6917629,08 2 The RMS value is then obtained using the last value from the column I Mot dt : I rms = 472 6917629 = 1518 . A < IU n 300 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 4 Information for specific applications 1200 1000 Torque in Nm 800 M per. M mot M load 600 400 200 0 0 200 400 600 800 1000 1200 1400 1600 Speed in RPM Load torque, motor torque and permissible torque as a function of the speed 300 Motor current in A 250 200 150 100 50 0 0 50 100 150 200 250 300 Time in s Motor current in a 300 s interval during acceleration and after acceleration Siemens AG SIMOVERT MASTERDRIVES - Application Manual 473 4 Information for specific applications 09.99 4.9 Buffering multi-motor drives with kinetic energy 4.9.1 General information In addition to buffering the system at power failure using additional capacitors in the DC link, it is also possible to buffer the system using the kinetic energy of the drives. For a multi-motor drive with common DC link bus, for stability reasons, only one drive can be used for the kinetic buffering. This drive must have sufficient kinetic energy, to cover the power requirement of the multi-motor drive during the power failure (power outage) time. During the buffering phase, the speed of the buffer drive drops, depending on the moment of inertia and the power requirement. If a specific speed ratio is required for the individual drives, then this must be ensured using the buffer drive as master drive. If, within the multi-motor group, there is no drive with sufficient kinetic energy, then an additional motor with coupled flywheel can be used as buffer drive. This motor is connected to the common DC link bus via an appropriate inverter. With this arrangement, the speed of the driven loads can be maintained during buffering, only the speed of the buffer drive decreases. The kinetic buffering function must be activated for all inverters. For the buffer drive inverter, the threshold when the kinetic buffering kicks-in, must be set as high as possible (e.g. 85%); for all of the other inverters, it should be set as low as possible (e.g. 65%). This prevents the drives from mutually influencing one another. The kinetic buffering is activated in the inverters of the remaining drives to prevent shutdown with DC link undervoltage. When dimensioning the system, it must be taken into account, that during buffering, the required power of the remaining drives including the losses, flows through the motor and the inverter of the buffer drive. For reasons of stability, the motor torque of the buffer drive should not exceed the rated torque. 474 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 4 Information for specific applications EE buffer drive INV 2 INV 1 INV P ( Pkin - Pload P ) mot P M M M P P load 1 load 2 P load P Power relationships during buffering Buffering using a drive in the multi-motor drive group For simplification, all drives are considered to be constant-torque drives. Only the moment of inertia of the buffer drive is taken into account. During buffering, the speed ratios should be maintained between the drives. The required power in the DC link during buffering is then obtained from: PDClink = i Pload max i Mot P Mot i WR Mot P max motoring The buffer drive feeds back into the DC link: PDClink = PMot P Mot P WR generating The following must be true: Pload max i i Mot i Mot P WR Mot P max = - PMot P Mot P WR Siemens AG SIMOVERT MASTERDRIVES - Application Manual 475 4 Information for specific applications 09.99 During buffering, the buffer drive must provide a variable-speed power of PMot P = - i Pload max i Mot i Mot P 2 WR Mot P Mot P max = - Perf max Mot P Mot P max The following is true: PMot P = M Mot P Mot P = ( J P d Mot P dt + M load P ) Mot P = - Pkin + Pload P A linear speed reduction is obtained with the simplifications/approximations which have been made. It is: d Mot P dt =- Perf max 1 ( + M load P ) J P Mot P max with Mot P = 2 n Mot P 60 buffer drive speed n Mot P : n Mot P max : buffer drive speed at the start of buffering JP : moment of inertia of the buffer drive including motor For the characteristics of the motor speed of the buffer drive with respect to time, during buffering, the following expression is obtained: n Mot P = n Mot P max - Perf max 60 ( + M Last P ) t 2 J P Mot P max The motor torque of the buffer drive should not be greater than the rated torque during buffering, i.e.: M Mot P M Mot P n With M Mot P = PMot P Mot P = Perf max Mot P max = constant the max. required power is given by the following: Perf max M Mot P n Mot P max 476 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 4 Information for specific applications Buffering via an additional buffer drive The speed of the driven load should, during buffering, be maintained for a specific time tP. The following is therefore true for the power required in the DC link: PDClink = i Pload i motoring Mot i WR The buffer drive feeds back into the DC link (regenerative power): PDClink = PMot P Mot P WR generating Thus, during buffering, the buffer drive must provide a constant power of PMot P = - i Pload i 2 Mot i Mot P WR = - Perf The energy is given by: WP = Wkin = 1 J ( 2Mot P max - 2Mot P min ) = Perf t P 2 P For the characteristics of the motor speed of the buffer drive over time, during buffering, the following expression is obtained: n Mot P = 2 Perf 60 2Mot P n - t 2 JP The motor torque of the buffer drive is given by M Mot P = Perf increases with decreasing speed Mot P At mot P=mot P min, it should be the same as the rated torque. If the buffer drive is only used up to the rated speed, i.e. mot P max=mot P n, then the required rated motor output is given by: PMot P n = M Mot P n Mot P n = Perf Mot P n Mot P min The required rated motor output therefore increases with mot P n/mot P min. Thus, for example, mot P n/mot P min is set to 1.1. The required quantities are obtained as follows: PMot P n = Perf 11 . J P = J Mot P + J sup pl = 2 Perf t P 2Mot P n (1 - 1 ) 11 .2 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 477 4 Information for specific applications 09.99 with Mot P n = 2 n Mot P n 60 As the mot P and nmot P n values, required to calculate Perf and JP, are still not known, a motor must be assumed. 478 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 4.9.2 4 Information for specific applications Example for buffering with an additional buffer drive A multi-motor drive, comprising 5 motors 30 kW/4-pole, 5 inverters 30 kW and a rectifier unit 160 kW, should continue to operate for a maximum of 1 second with the same motor speeds when power failures occur. The power requirement is 150 kW. The required energy during buffering, including the losses, should be covered by the kinetic energy of a high-inertia flywheel load. In this case an additional inverter with induction motor and coupled flywheel should be connected. With this arrangement, the speed of the driven loads can be maintained during buffering; only the buffer drive speed decreases (flywheel drive). The following configuration is obtained. EE 160 kW buffer drive INV IM 30 kW 30 kW INV IM 30 kW 30 kW INV IM 5x When the system is powered-up, at first, the induction motor with the coupled flywheel is accelerated. The driven loads are then powered-up. This prevents the rectifier unit from being overloaded. In operation, the buffer drive practically only draws the reactive power. The DC link current of the associated inverter is therefore 0, as only the losses of the motor, running under no-load conditions, and the inverter losses have to be covered. When a power failure occurs, the buffer drive supplies the required power to the DC link. Siemens AG SIMOVERT MASTERDRIVES - Application Manual 479 4 Information for specific applications 09.99 Dimensioning the buffer drive For the given multi-motor drive, with the following data Pload = 30 kW (the same load for every motor), Mot = 0.918 the following dimensioning data are obtained for the buffer drive. In this case, a 2-pole motor is to be used. The following assumption is made: A 200 kW motor with mot P=0.959 and nmot P n=2980 RPM. The inverter efficiency is assumed to be INV=0.98. The following is therefore obtained: Perf = i Pload i Mot i Mot P 2 WR = 5 30 = 177.4 kW 0.918 0.959 0.982 PMot P n = Perf 11 . = 177.4 12 . = 195.2 kW Mot P n = JP = 2 n Mot P n 60 = 2 Perf t P 2Mot P n (1 - 1 ) .2 11 2 2980 = 312.06 s -1 60 = 2 177.4 10 3 1 = 21 kgm2 1 312.062 (1 - 2 ) 11 . The minimum motor speed of the buffer drive is given by: n Mot P min = n Mot P n 11 = 2980 = 2709 RPM 11 . Thus, a 200 kW motor 1LA6 317-2AC.. and a 200 kW inverter 6SE7033-7TG60 with vector control is therefore used for the buffer drive. The additional moment of inertia is then given by: J sup pl = J P - J Mot = 21 - 2.3 = 18.7 kgm 2 The calculated additional moment of inertia can, for example, be implemented using a steel flywheel. The following is valid: J= d 7.85 r 4 1000 2 [kgm2] d, r in m The following is obtained for r if d is set to 0.1 m: r=4 480 2 J 2 18.7 =4 = 0.351 m 3 . 7.85 103 d 10 01 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 4 Information for specific applications The diameter of the 10 cm thick flywheel is therefore 70.2 cm. The additional 200 kW inverter in conjunction with the 160 kW rectifier unit is permissible, as, in operation, the buffer drive practically requires no active power, and, additionally, the following condition is fulfilled: PEE 0.3 Pinv. = 0.3 (5 30 + 200) = 105 kW In order not to overload the rectifier unit while the flywheel drive is accelerating, the acceleration time is, for example, adjusted, so that when the rated speed is reached, precisely 160 kW motor output is required (=rated output of the rectifier unit). The motor output is given by: PMot P max = M b Mot P n = J P Mot P n 2 n Mot P n 2 1 Mot P n = J P ( ) tH 60 tH or for the acceleration time: tH = JP ( 2 n Mot P n 60 )2 1 PMot P max = 21 ( 2 2980 2 1 ) = 12.8 s 60 160 103 As the acceleration time in this case, is less than 60 s, a shorter acceleration time can be achieved by utilizing the overload capability of the rectifier unit. Checking the inverter rating For the multi-motor drive inverters, it must be observed, that the connected motors must go into the field-weakening mode earlier, due to the fact that the DC link voltage is reduced to 85% in buffer operation. This means that a higher current is drawn. Depending on the pre-loading condition and duration of the buffer time, this can result in an inverter overload. The maximum motor current in field-weakening operation is given by: I Mot max = I Mot n ( kn = I mag 2 I mag 2 1 M load 2 ) (1 - ( ) ) kn 2 + ( ) 2 M Mot n I Mot n I Mot n kn n Mot max 0.85 n Mot n Siemens AG SIMOVERT MASTERDRIVES - Application Manual 481 4 Information for specific applications 09.99 In this particular case, the 30 kW motors are operated with rated torque and rated speed. Then the following is true: M load =1 M Mot n kn = n Mot n 0.85 n Mot n = 1 0.85 With I mag I Mot n = 0.41 I Mot n = 55 A the maximum motor current is given by: I Mot max = 62 A The RMS value in a 300 s interval is then calculated as follows: I eff = 2 2 I Mot max t P + I Mot n ( 300 - t P ) 300 = 62 2 1 + 552 (300 - 1) = 55.02 A 300 The 30 kW inverter, with a rated current of 59 A and a maximum current of 80.2 A, are therefore adequately dimensioned. For the inverter of the buffer drive, there are practically no restrictions due to the reduced DC link voltage, as in this case, there is no pre-loading, and because, at nmot P max, the motor current increase as a result of the field-weakening factor kn=1/0.85, is partially compensated by the low motor torque (Mmot P=Mmot P n/1.1). 482 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 4 Information for specific applications 4.10 Harmonics fed back into the supply in regenerative operation Presently, harmonics fed back into the supply can only be calculated using the PATH configuring program, for 1Q operation. Thus, the basic differences between motoring using an uncontrolled rectifier and generating using a rectifier/regenerative feedback unit are to be investigated here. The following equivalent circuit diagrams are used as basis for these investigations. ~ i supply u i DC link ~ C DC link Iload u DC link ~ Usupply Lsupply 6-pulse uncontrolled rectifier 3 Equivalent circuit diagram when motoring with an uncontrolled rectifier i supply u I load i DC link ~ ~ C DC link u DC link ~ U supply 3 L supply Autotransformer 1:1,2 6-pulse controlled rectifier Equivalent circuit diagram when generating with a rectifier/regenerative feedback unit Siemens AG SIMOVERT MASTERDRIVES - Application Manual 483 4 Information for specific applications 09.99 The load current Iload is assumed to be constant. It is calculated as follows: I load = PMot U DC link Mot INV motoring I load = PMot Mot INV U DC link generating The approximate DC link voltage is obtained, when motoring, without taking into account the voltage drops as follows: U DC link 1.35 U sup ply When generating with the rectifier/regenerative feedback unit, the DC link voltage is specified through the closed-loop control. The DC link capacitor CDC link is provided through the inverter or drive converter. The inductance Lsupply is calculated from the fault level on the high voltage side SK HS, the nominal transformer rating STr n, the uk of the transformer, the uk of the commutating reactor and the rated apparent power of the inverter or inverter SU n. For regenerative operation with the rectifier/regenerative feedback unit, the autotransformer u k is also included. The total inductance Lsupply is given by: Lsup ply = LK HS + LK Tr + LK Dr + ( LK SpTr ) with K HS L = LK Tr = LK Dr = 2 U sup ply S K HS 2 U sup ply u k Tr STr n 2 U sup ply u k Dr LK SpTr = SU n 2 U sup ply u k SpTr SU n component of the high voltage network component of the converter transformer component of the reactor component of the autotransformer (for rectifier/regenerative feedback unit) = 2 f 484 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 4 Information for specific applications When calculating using an Excel/VBA program, the characteristics of the line current over 3 periods of the line supply voltage are calculated, starting from practical initial values. The steadystate condition is then achieved. The 4th period is therefore output, and Fourier analysis applied. Depending on the load and line supply inductance, operation with continuous current or discontinuous current is obtained. When motoring using an uncontrolled rectifier, the DC link voltage is obtained according to the line supply voltage and load. The load current must be determined iteratively. When generating with a rectifier/regenerative feedback unit, the DC link voltage is controlled (closed-loop). The firing angle St is therefore iteratively determined corresponding to the control input for the DC link voltage. Calculation example For a drive with a 90 kW load, the relationships regarding the harmonics fed back into the line supply are to be investigated, when motoring/generating. When motoring, a 90 kW drive converter with 2% uk line reactor is assumed, and for generating, a 90 kW inverter with 90 kW rectifier/regenerative feedback unit and 4% uk line reactor. A uk of 2% is assumed for the autotransformer. The data regarding supply voltage (2.2 kV/400 V), transformer (STr=150 kVA, uk Tr=4%) and line supply fault level (SK HS=10 MVA) are the same. The motor efficiency is 94.9% and the inverter efficiency is assumed to be 98%. The DC link capacitance of the drive converter and inverter is 12 mF. The inductance Lsupply is given by L 400 2 = = 51 H 10 10 6 314 LK Tr 400 2 0.04 = = 136 H 150 10 3 314 K HS LK Dr 2% = 400 2 0.02 = 79 H 3 400 186 314 (with IU n=186 A) LK Dr 4% = 158 H LK SpTr 2% = 79 H to Lsup ply mot = LK HS + LK Tr + LK Dr 2% = 266H Lsup ply gen = LK HS + LK Tr + LK Dr 4% + LK SpTr 2% = 424H Siemens AG SIMOVERT MASTERDRIVES - Application Manual 485 4 Information for specific applications 09.99 The calculation results in a 521 V DC link voltage when motoring. The load current is then given by I load 90 10 3 PMot = = 186 A = U DC link Mot INV 521 0.949 0.98 The same DC link voltage is assumed when generating. The load current is then obtained as follows I load = PMot Mot INV 90 103 0.949 0.98 = = 161 A 521 U DC link Additional results When motoring: I sup ply(1) = 145.5 A basic line current fundamental, secondary side cos (1) = 0.963 power factor of the basic fundamental I sup ply RMS = 153.2 A RMS line current, secondary side DFsec = 7.2% distortion factor on the secondary side DFprim = 2% distortion factor on the primary side Generating operation: I sup ply (1) = 152 A basic fundamental of the line current, secondary side cos (1) = 0.796 power factor of the basic fundamental I sup ply RMS = 162.5 A RMS line current, secondary side DFsec = 8.6% distortion factor on the secondary side DFprim = 2.4% distortion factor on the primary side St = 1413 . 0 rectifier firing angle 486 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 4 Information for specific applications 400 Current in A, voltage in V 300 200 100 i supply u u supply u 0 0 0.002 0.004 0.006 0.008 0.01 0.012 0.014 0.016 0.018 0.02 -100 -200 -300 -400 Time in s Line current and associated phase voltage when motoring 400 Current in A, voltage in V 300 200 100 i supply u u supply u 0 0 0.002 0.004 0.006 0.008 0.01 0.012 0.014 0.016 0.018 0.02 -100 -200 -300 -400 Time in s Line current and associated phase voltage when generating Siemens AG SIMOVERT MASTERDRIVES - Application Manual 487 4 Information for specific applications 09.99 900 Voltage in V, current in A 800 700 600 500 u DC link i DC link 400 300 200 100 0 0 0.002 0.004 0.006 0.008 0.01 0.012 0.014 0.016 0.018 0.02 0.018 0.02 Time in s DC link voltage and DC link current when motoring 300 200 Voltage in V, current in A 100 0 0 0.002 0.004 0.006 0.008 0.01 0.012 0.014 0.016 -100 u DC link i DC link -200 -300 -400 -500 -600 Time in s DC link voltage and DC link current when generating 488 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 4 Information for specific applications 100 90 80 70 60 50 40 30 20 10 0 1 5 7 11 13 17 19 23 25 Line harmonic currents as a % of the basic fundamental when motoring 100 90 80 70 60 50 40 30 20 10 0 1 5 7 11 13 17 19 23 25 Line harmonic currents as a % of the basic fundamental when generating Siemens AG SIMOVERT MASTERDRIVES - Application Manual 489 4 Information for specific applications 09.99 600 Current in A, voltage in V 500 400 u DC link i DC link 300 200 100 0 0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 Time in s DC link current and DC link voltage stabilizing over three line supply periods when motoring Summary When generating, the load current in the DC link is lower than when motoring, due to the influence of the efficiencies. However, a higher line current Isupply (1) is obtained than when motoring due to the fact that when generating, the cos (1) is poorer (the firing angle is not 0), and due to the ratio of the autotransformer of 1.2 (the current is transformed up). Inspite of the higher line inductance, higher harmonic currents are obtained due to the unfavorable firing angle of the rectifier when generating. 490 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 4 Information for specific applications 4.11 Calculating the braking energy of a mechanical brake 4.11.1 General information Many drives are equipped with a mechanical holding brake. At standstill with the motor shutdown, this brake is intended to prevent any motion. This is extremely important for drives with a significant holding torque, for example, hoisting and elevator drives. When the electrical braking function fails, the mechanical holding brake must also assume an emergency stop function. Two examples to calculate the braking energy are now shown. As a result of supplementary torques (hoisting torque and frictional torque), with these examples, it is not possible to simply apply the condition "max. external moment of inertia = motor intrinsic moment of inertia". 4.11.2 Hoisting drive with counterweight The braking energy is to be calculated for the most unfavorable case for a mechanical brake for a hoisting drive with counterweight. It has been assumed that the electrical braking has failed. Brake M D Gearbox F+Q v down ~ ~ ~ ~ Counterweight Chain Block diagram of the hoisting drive Siemens AG SIMOVERT MASTERDRIVES - Application Manual 491 4 Information for specific applications 09.99 Drive data Weight of the elevator cabin mF = 2000 kg Weight of the load mQ = 5100 kg Weight of the counterweight mG = 4500 kg Weight of the chain mchain = 200 kg Moment of inertia, motor JMot = 0.36 kgm2 Moment of inertia, gearbox Jgearbox = 0.012 kgm2 Moment of inertia, brake Jbrake = 0.02 kgm2 Gearbox ratio i = 16.79 Pulley diameter D = 0.26 m Max. traversing velocity vmax = 1.3 m/s Braking torque of the mechanical brake Mbr = 400 Nm Permissible braking energy of the mechanical brake W br max = 100 kJ Mechanical hoisting drive efficiency mech = 0.9 Gearbox efficiency gearbox = 0.95 The worst case is motion downwards at full velocity and with a full load. In this case, there is the highest kinetic energy and the mass difference acts to accelerate the cabin downwards. The braking energy can be calculated using the energy balance equation. The following is true: Wbr = Wkin mot + Z + mech gearbox (Wkin load + W pot load ) with Wkin mot + Z = Wkin load = 1 ( J mot + J brake + J gear ) ( mot 2 max )2 = 2 v max 2 1 ) J mot + Z (i D 2 1 2 m v max 2 W pot load = m g h Further, h = v max 492 t br 2 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 4 Information for specific applications Wbr = M br mot m = m F max 2 v max t br t br = M br i 2 2 D + mQ + mG + mchain m = mF + mQ - mG Appropriately combined, the following is obtained: 1 2 vmax 2 1 2 ) + mech gear m vmax J mot + Z (i D 2 Wbr = 2 D mech gear m g 2 1- M br i From this formula it can be seen that braking can only function, if: M br > mech gear m g D 2 i The steady-state load torque, referred to the motor speed must therefore be lower than the braking torque of the mechanical brake which is available. The following is obtained with the specified numbers: m = 2000 + 5100 + 4500 + 200 = 11800 kg m = 2000 + 5100 - 4500 = 2600 kg J Mot + Z = 0.36 + 0.02 + 0.012 = 0.392 kgm2 . 2 1 2 13 1 0.392 (16.79 ) + 0.9 0.95 11800 13 . 2 . 2 0 26 2 Wbr = = 24314 Ws = 24.314 kJ 0.26 0.9 0.95 2600 9.81 2 1- 400 16.79 t br = Wbr D 24314 0.26 = = 0.724 s M br i v max 400 16.79 13 . Thus, the braking energy which occurs, is less than the permissible braking energy of the mechanical brake. Siemens AG SIMOVERT MASTERDRIVES - Application Manual 493 4 Information for specific applications 09.99 The braking energy can also be calculated by analyzing the torques. The braking torque when moving downwards is as follows: M br = J mot + Z i load + ( J load load + M load ) mech gearbox i with load = load max 2 v max = t br D tbr D J load = m ( ) 2 2 M load = m g D 2 Further, Wbr = M br mot J max 2 v max t br t br = M br i 2 2 D = J mot + J brake + J gear mot + Z m = m F + mQ + mG + mchain m = mF + mQ - mG The same formula as for W br is obtained by appropriately inserting. 494 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 4.11.3 4 Information for specific applications Traversing drive For a traversing drive at maximum velocity, the braking energy occurring at the mechanical brake is to be calculated. It is assumed that the electrical braking has failed. v Guide m Brake M D Gearbox Block diagram of the traversing drive Drive data Weight of the load m = 5500 kg Moment of inertia, motor Jmot = 0.017 kgm Moment of inertia, gearbox Jgearbox = 0.00164 kgm2 Moment of inertia, brake Jbrake = 0.00063 kgm2 Gear ratio i = 16 Wheel diameter D = 0.34 m Max. traversing velocity vmax = 2.66 m/s Braking torque of the mechanical brake Mbr = 60 Nm Permissible braking energy of the mechanical brake W br max = 25 kJ Traversing resistance wF = 0.02 Gearbox efficiency gearbox = 0.95 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 2 495 4 Information for specific applications 09.99 The braking energy is calculated using the energy equation. The following is valid: Wbr = Wkin mot + Z + (Wkin load - W friction ) VZ gear with VZ = sign(Wkin load - W friction ) Wkin mot + Z = Wkin load = 1 ( J mot + J brake + J gear ) ( mot 2 (normally positive) max )2 = 2 v max 2 1 ) J mot + Z (i D 2 1 2 m v max 2 W friction = m g wF s Further, s = v max t br 2 Wbr = M br mot max 2 v max t br tbr = M br i 2 2 D Re-arranging, the following is obtained: 1 2 vmax 2 1 2 ) + VZ J mot + Z (i m vmax gear 2 D 2 Wbr = D VZ gear m g wF 2 1+ M br i To start off with, the sign of the gearbox efficiency must be determined. The limiting case with W kin load=W friction is investigated. The following is obtained for Wbr: Wbr = 1 2 vmax 2 2 vmax tbr J mot + Z (i ) = M br i 2 D 2 D Further, with W kin load=W friction, the following is obtained t 1 2 m v max = m g wF v max br 2 2 496 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 4 Information for specific applications After eliminating tbr, an equation is obtained for wF: wF = D M br J mot + Z i 2 g If the following is now valid wF D M br J mot + Z i 2 g -1 the sign should be set positive. Otherwise, set the sign negative ( =1/). The following is obtained with the specified numerical values: J mot + Z = 0.017 + 0.00063 + 0.00164 = 0.01927kgm 2 0.34 60 D M br = = 3.372 > 0.02 J mot + Z i 2 g 0.01927 16 2 9.81 (i. e. the sign is positive) 1 2 2.66 2 1 0.01927 (16 ) + 0.95 5500 2.66 2 0.34 2 Wbr = 2 = 16156 Ws = 16156 kJ . 0.34 0.95 5500 9.81 0.02 2 1+ 60 16 t br = Wbr D 16156 0.34 = . s = 2151 M br i v max 60 16 2.66 The braking energy obtained is therefore less than the permissible braking energy of the mechanical brake. Siemens AG SIMOVERT MASTERDRIVES - Application Manual 497 4 Information for specific applications 09.99 The braking energy can also be calculated by investigating the torques. The following is valid for the braking torque: M br = J mot + Z i load + ( J load load - M friction ) VZ grearbox i with VZ = sign( J load load - M friction ) load = load max 2 v max = t br D tbr D J load = m ( ) 2 2 J mot + Z = J mot + J brake + J gearbox M friction = m g wF D 2 Further, Wbr = M br mot max 2 v max t br t br = M br i 2 2 D The same formula is obtained for W br by appropriately applying. 498 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 4 Information for specific applications 4.12 Criteria for selecting motors for clocked drives 4.12.1 General information Clocked drives with 1FK6/1FT6 synchronous servomotors or 1PA6 induction servomotors are to be investigated. These can include, for example, traversing drives, rotary table drives and spindle drives. The motors must fulfill, on one hand the dynamic requirements when accelerating and decelerating, and on the other hand, operation must be permissible from a thermal perspective. A drive task can generally be solved using several motors with various rated speeds. The rated speed is selected according to the following criteria: * minimum motor size * minimum motor current (drive converter selection) * preferred motor series, e. g. 1FK6 with nn=3000 RPM and nn=6000 RPM * existing motor outputs at a specific rated speed A gearbox is generally used to adapt the load torque and load speed. In this case, a differentiation must be made between wheels: * a gearbox ratio is specified (e. g. toothed belt ratio) * the gearbox ratio can be selected (e. g. planetary gearbox for motor mounting) For the latter, an additional degree of freedom is obtained when selecting the motor. For example, the optimum gearbox ratio for a minimum motor torque can be defined. However, the gearbox ratio which can be selected is restricted by the maximum permissible motor speed, as well as the possible ratios for certain gearboxes. Other restrictions are obtained due to the size, efficiency and costs. These must be taken into account when finally selecting the drive package, comprising motor, gearbox and drive converter. Basic considerations regarding motor utilization Optimum motor utilization according to the dynamic requirements In order to determine the optimum dynamic utilization of a motor, the dynamic limiting curve of the motor is investigated with the resulting permissible motor output. This is given by: Pmot perm dyn = M mot perm dyn 2 nmot 60 Siemens AG SIMOVERT MASTERDRIVES - Application Manual (nmot in RPM) 499 4 Information for specific applications 09.99 2.5 2 1.5 M perm. / M perm. max P perm. / P rated n limit 1 0.5 n rated 0 0 500 1000 1500 2000 2500 3000 3500 4000 4500 Motor speed in RPM Example for the dynamic limiting curve and the dynamic permissible motor output for 1FT6082-8AF7 2.5 2 1.5 M perm. / M perm. max P perm. / P rated n limit 1 0.5 n rated 0 0 500 1000 1500 2000 2500 3000 3500 4000 Motor speed in RPM Example for the dynamic limiting curve and the dynamic permissible motor output for 1PA6103-4HG M perm. = M perm.max = 2 M mot n for nmot nlim it , 500 M perm. = M stall nmot n 2 ( ) 1.3 nmot for n mot > n lim it Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 4 Information for specific applications A motor is dynamically suitable for a particular application if all of the torque/speed points lie below the dynamic limiting curve. The maximum permissible motor torque Mperm. max can only be utilized up to the limiting speed nlimit. After this, there is a decrease due to the voltage limiting curve and the stall torque limiting curve. However, for synchronous servomotors, the maximum possible motor output is generally achieved for a motor speed, which lies somewhat higher than n limit, but is less than nrated. For induction servomotors, the maximum possible motor output is always reached at nlimit, as the stall torque limit characteristic is proportional to 1/nmot2. A pure flywheel drive (high-inertia drive) with constant acceleration is now to be investigated. For the specified max. load speed nload max and when using a gearbox with moment of inertia JG, the maximum motor torque and the maximum motor output when accelerating are given by: M mot max = ( J mot + J G ) load max i + J load load max Pmot max = M mot max 1 i 2 nmot max (nmot max in RPM) 60 = Pload max + ( J mot + J G ) load max i 2 nmot max 60 with i= nmot max nload max Pload max = J load max load max = J load 2 nload max 2 nload max 60 (nload max in RPM) 60 M mot max ( J mot + J G ) i + J load 1 i The following curves are obtained, if the points Pmot max/nmot max and Pload max/nmot max are entered for a constant Mmot max for various values of i. Siemens AG SIMOVERT MASTERDRIVES - Application Manual 501 4 Information for specific applications 09.99 P load max P mot max i n mot max Example for the max. motor output and the max. load power as a function of the gearbox ratio i The optimum gearbox ratio for the highest power output to the load, and therefore for the shortest accelerating time, is given by dPload max di = 0 = - J load 2 nload max 60 M mot max 1 (( J mot + J G ) i + J load ) 2 i ( J mot + J G - J load 1 ) i2 to iopt = J load J mot + J G If the acceleration load max is specified, with the optimum gearbox ratio, the lowest accelerating torque Mmot max can be achieved, and therefore also the lowest motor current. The following is obtained for the max. motor output with i=iopt: Pmot max = 2 Pload max 502 (i. e. power adaption) Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 4 Information for specific applications From the output perspective, a motor is best utilized when it comes to the dynamic limiting characteristic, if the required max. motor output lies close to the max. possible motor output. The max. load power is reached at a constant max. motor torque using the optimum gearbox ratio. If the max. motor speed, achieved with the optimum gearbox ratio, is higher than nlimit, then the maximum of the load output is obtained at nlimit. Depending on the ratio of the motor moment of inertia to the load moment of inertia, a differentiation can be made between 2 limiting cases. a) The motor moment of inertia has no significant influence on the acceleration For all of the gearbox ratios coming into question the following is valid: J mot + J G << J load / i 2 As far as the Pload max/nmot max characteristic is concerned, nmot max lies to the far left away from the optimum position. The following is obtained for Pload max: Pload max Pmot max The required output Pload max can be achieved with a high torque at a low speed, or with a low torque at a high speed. As far as the dimensions are concerned, generally, motors with a higher rated speed are more favorable, as the maximum possible motor output increases, for the same dimensions, with higher rated speeds. The utilization is at an optimum, if the motor is being precisely operated at the maximum possible motor output. b) The motor moment of inertia has a significant influence on the acceleration In this case, it is favorable to set the optimum gearbox ratio. For i = iopt , the following is obtained for Pload max: Pload max = Pmot max 2 Also here, as far as the dimensions are concerned, generally motors with higher rated speeds are more favorable. The motor utilization is at an optimum, if the motor is operated at nlimit with the maximum motor output possible at that point. Siemens AG SIMOVERT MASTERDRIVES - Application Manual 503 4 Information for specific applications 09.99 Optimum utilization of a motor according to the thermal requirements In order to determine the optimum thermal utilization of a motor, the S1 limiting characteristic of the motor is investigated and the resulting permissible motor output. This is given by: Pmot perm.stat = M mot perm.stat 2 nmot 60 (nmot in RPM) 1.2 1 0.8 M perm. S1 / M 0 P perm. / P rated 0.6 0.4 0.2 n rated 0 0 500 1000 1500 2000 2500 3000 3500 4000 4500 Motor speed in RPM Example for the S1 limiting curve and the steady-state permissible motor output for 1FT6086-8AH7 504 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 4 Information for specific applications 1 0.9 n nenn 0.8 0.7 0.6 M perm. S1 / M rated P perm. / P rated 0.5 0.4 0.3 0.2 0.1 0 0 500 1000 1500 2000 2500 3000 3500 4000 Motor speed in RPM Example for the S1 limiting curve and the permissible steady-state motor output for 1PA6103-4HG A motor is thermally suitable for a particular drive situation, if the RMS motor torque at naverage lies below the S1 limiting characteristic. For synchronous servomotors, the thermally permissible torque decreases with increasing motor speed. The maximum possible motor output is however reached for a motor speed which generally lies above the rated speed. For synchronous servomotors with higher rated speeds and higher rated outputs, this value can also lie below the rated speed. For induction servomotors, the thermally permissible torque for motor speeds greater than nrated linearly decreases with 1/nmot. Thus, the maximum possible motor output is achieved at nrated and then remains constant. Thus, a motor would be well utilized, as far as the S1 limiting characteristic is concerned, if the average motor speed is close to the power optimum. However, for clocked drives, this is generally not possible due to a reduction in the permissible dynamic motor torque at nmot max due to the voltage limiting characteristic and the stall torque limiting characteristic. Thus, if a drive with a constant accelerating torque is investigated, the average motor speed may only be so high, that at nmot max, the accelerating torque is just below the dynamic limiting characteristic. The optimum utilization is achieved, if, in addition, the RMS motor torque is close to the S1 limiting characteristic and if the average motor speed is just reached with the optimum gearbox ratio for the lowest RMS motor torque (for a pure flywheel drive [high-inertia drive]), the minimum of the RMS motor torque coincides with the minimum of the accelerating torque). Of course, these conditions can only seldomly be simultaneously fulfilled. Siemens AG SIMOVERT MASTERDRIVES - Application Manual 505 4 Information for specific applications 09.99 Procedure when selecting the rated speed for a specific drive task a) Gearbox ratio is specified * motors with various rated speeds are determined using the PATH configuring program * a suitable motor/drive converter combination is selected b) The gearbox ratio can be selected * a motor with the smallest frame size for a specific rated speed is determined using the PATH configuring program, by varying the gearbox ratio within the possible limits * this technique is then applied for other rated speeds * a suitable motor/gearbox/drive converter combination is selected Depending on whether the motor is selected according to the RMS value (e. g. short no-load intervals), according to the maximum motor torque (e. g. long no-load intervals) or additionally according to the optimum gearbox ratio, for a specific drive task, a motor with lower or higher rated speed might be more favorable. The following examples will clearly illustrate this. 506 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 4 Information for specific applications 4.12.2 Example 1, rotary table drive with i=4 and short no-load interval Drive data 2 Load moment of inertia Jload = 0.4 kgm Moment of inertia of the gearbox (toothed belt pulleys) JG = 0.002 kgm Angle tot. = 180 degrees Positioning time ttot. = 0.2 s No-load interval tp = 0.2 s 2 A 1FT6 motor is to be used. Traversing characteristic load load max tot. t t tot. load max = load max = nload max = tp 2 tot . 2 = = 31.42s -1 ttot . 0 .2 load max 2 31.42 = = 314.2s -2 ttot . 0.2 2 load max 60 31.42 60 = = 300RPM 2 2 nmot max = i nload max = 4 300 = 1200 RPM Siemens AG SIMOVERT MASTERDRIVES - Application Manual 507 4 Information for specific applications 09.99 Thus, the maximum motor torque when accelerating or decelerating is given by: M mot max = ( J mot + J G ) load max i + J load load max 1 i The RMS motor torque and the average motor speed are given by: M rms = naverage 2 M mot max ttot . ttot . + t p = M mot max ttot . 0.2 = M mot max = M mot max 0.707 ttot . + t p 0.2 + 0.2 nmot max ttot . 1200 2 0 .2 2 = 2 = 2 = 300 RPM ttot . + t p 0 .2 + 0 .2 For the specified gearbox, the maximum motor torque and the RMS motor torque only depend on Jmot. When selecting an 1FT6 motor, the following 3 motors are obtained Rated speed RMS motor current Motor type / rated output Drive converter / rated output 1500 RPM 13.09 A 1FT6105-8AB7 / 6.6 kW 6SE7021-4EP50 / 5.5 kW 2000 RPM 17.45 A 1FT6105-8AC7 / 8 kW 6SE7022-1EP50 / 7.5 kW 3000 RPM 26.18 A 1FT6105-8AF7 / 9.7 kW 6SE7022-7EP50 / 11 kW Due to the short no-load interval, the motor is selected according to the RMS value. The 3 motors have the same frame size and length, have different rated speeds and rated outputs, but the same motor moment of inertia and the same S1 characteristic. In this case, the most favorable motor is that with the lowest motor current, i. e. in this case, the motor whose rated speed is located close to the average motor speed naverage=300 RPM. The motor type with 1500 RPM fulfills this condition the best, and the smallest converter, with 5.5 kW is obtained. 508 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 4 Information for specific applications 50 45 40 Torque in Nm 35 30 25 M perm. S1 M RMS / n average 20 15 10 5 0 0 500 1000 1500 2000 2500 3000 Motor speed in RPM MS1 characteristics and MRMS at naverage for motors with nn=1500 RPM, nn=2000 RPM and nn=3000 RPM 140 M mot, M perm. in Nm 120 100 80 60 40 20 0 0 200 400 600 800 1000 1200 1400 1600 Motor speed in RPM Dynamic limiting curve and motor torque characteristic for the motor with nn=1500 RPM Siemens AG SIMOVERT MASTERDRIVES - Application Manual 509 4 Information for specific applications 4.12.3 09.99 Example 2, rotary table drive with i=4 and long no-load interval The rotary table drive from example 1, with otherwise identical data, but a no-load interval of 4 s is now to be investigated. The average motor speed is: naverage nmot max ttot . 1200 2 0 .2 2 = 2 = 2 = 28.57RPM ttot . + t p 0 .2 + 4 The ratio of the RMS torque to the maximum torque is given by: M rms = M mot max 0.2 = M mot max 0.22 0.2 + 4 The following 4 1FT6 motors are selected Rated speed Max. motor current Motor type / rated output Drive converter / rated output 2000 RPM 19.52 A 1FT6084-8AC7 / 3.5 kW 6SE7018-0EP50 / 3 kW 3000 RPM 29.42 A 1FT6064-6AF7 / 2.2 kW 6SE7021-0EP50 / 4 kW 4500 RPM 44.15 A 1FT6064-6AH7 / 2.3 kW 6SE7022-1EP50 / 7.5 kW 6000 RPM 62.15 A 1FT6082-8AK7 / 3.5 kW 6SE7022-7EP50 / 11 kW As a result of the longer no-load time, in this case the motor is selected according to the dynamic limiting characteristic. The selected drive converter, type of construction Compact Plus, is utilized for a 300 % rated current. This is possible, as the current flows for less than 250 ms and the noload time is greater than 750 ms. From the two smallest motors with nn=3000 RPM and nn=4500 RPM, that motor with the lowest rated speed is more favorable, as in this case the motor torque characteristic can be better adapted to the dynamic limiting characteristic, thus resulting in a lower motor current. Although the motor with nn=2000 RPM is larger, it has a lower motor current. The reason for this is that the maximum speed is closer to the rated speed. The optimum combination is the motor with nn=3000 RPM together with the 4 kW drive converter. 510 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 4 Information for specific applications 40 35 M mot, M perm. in Nm 30 25 M perm. (3000 RPM) M perm. (4500 RPM) M max 20 15 10 5 0 0 500 1000 1500 2000 2500 3000 3500 4000 4500 Motor speed in RPM Dynamic limiting characteristic and motor torque characteristic for motors with nn=3000 RPM and nn=4500 RPM 10 9 8 Torque in Nm 7 6 M perm. S1 M RMS / n average 5 4 3 2 1 0 0 500 1000 1500 2000 2500 3000 Motor speed in RPM MS1 characteristic and Mrms at naverage for the motor with nn=3000 RPM Siemens AG SIMOVERT MASTERDRIVES - Application Manual 511 4 Information for specific applications 4.12.4 09.99 Example 3, rotary table drive with a selectable ratio and short no-load interval Drive data Load moment of inertia Jload = 0.075 kgm2 Moment of inertia, gearbox JG = 0.0003 kgm2 Angle tot. = 18 degrees Positioning time ttot. = 0.05 s No-load time tp = 0.05 s A 1FT6 motor is to be used. load max 2 2 tot . = = 10 = 12.566s -1 ttot . 0.05 load max = nload max = load max 2 12.566 = = 502.65s - 2 ttot . 0.05 2 load max 60 12.566 60 = = 120 RPM 2 2 nmot max = i nload max The maximum motor torque when accelerating and when decelerating as well as the RMS motor torque is obtained, as in example 1, as follows: M mot max = ( J mot + J G ) load max i + J load load max M RMS = 2 M mot max t tot . t tot . + t p = M mot max 1 i t tot . 0.05 = M mot max = M mot max 0.707 t tot . + t p 0.05 + 0.05 For a selectable gearbox ratio, the maximum motor torque and the RMS motor torque are dependent on Jmot, JG and i. The average motor speed is given by: naverage 512 i nload max ttot . 2 2 2 = ttot . + t p Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 4 Information for specific applications 3 1FT6 motors are selected, taking into account the optimum gearbox ratio. Rated speed RMS motor current Motor type / rated output Drive converter / rated output 2000 RPM 3.44 A 1FT6064-6AC7 / 1.7 kW 6SE7015-0EP50 / 1.5 kW 3000 RPM 4.98 A 1FT6064-6AF7 / 2.2 kW 6SE7015-0EP50 / 1.5 kW 4500 RPM 7.51 A 1FT6064-6AH7 / 2.3 kW 6SE7018-0EP50 / 3 kW As a result of the short no-load time, the motor is selected according to the RMS value. The 3 motors have the same frame size and length, but have different rated speeds and rated outputs, and the same motor moment of inertia and the same S1 characteristic. Thus, when the gearbox is optimized, the same value is obtained for all of the 3 motors iopt = J load = J mot + J G 0.075 = 6.85 0.0013 + 0.0003 In the table above, 7 was used as the basis. The most favorable motor is that motor with the lowest motor current, i. e. the motor whose rated speed is the closest to the average motor speed. The motor type with 2000 RPM fulfills this condition. The motor with nn=1500 RPM cannot be used, as the smallest motor with this rated speed is the 1FT6102-8AB7 with a rated output of 3.8 kW. The maximum speed and the average motor speed are given by: n mot max = i nload max = 7 120 = 840 RPM naverage i nload max t tot . 7 120 2 0.05 2 2 2 = = = 210 RPM t tot . + t p 0.05 + 0.05 The following diagram is obtained if the points Mrms, naverage are entered into diagram Mperm. S1 for the various gearbox ratios. Siemens AG SIMOVERT MASTERDRIVES - Application Manual 513 4 Information for specific applications 09.99 12 i=3 M RMS, M perm. S1 in Nm 10 i=13 8 i=7 6 M perm. S1 M RMS / n average 4 2 0 0 200 400 600 800 1000 1200 1400 1600 1800 2000 n mittel in 1/min Optimizing MRMS for the motor with nn=2000 RPM If the transition points Mmot max, nmot max are entered into the diagram Mperm. max, for the various gearbox ratios, the following diagram is obtained. M mot max, M perm. in Nm 40 35 30 25 20 15 i=3 i=13 10 i=7 5 0 0 200 400 600 800 1000 1200 1400 1600 1800 2000 n mot max in RPM Optimizing Mmot max for a motor with nn=2000 RPM 514 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 4.12.5 4 Information for specific applications Example 4, rotary table drive with selectable ratio and long no-load interval The rotary table drive from example 1, with otherwise the same data but a no-load time of 4 s is now to be investigated with a selectable ratio (planetary gear). The appropriate gearbox moment of inertia is used depending on the selected motor type. The following 4 1FT6 motors were selected Rated speed Max. motor current Motor type / rated output Drive converter / rated output 2000 RPM 16.33 A 1FT6084-8AC7 / 3.5 kW 6SE7018-0EP50 / 3 kW 3000 RPM 14.55 A 1FT6064-6AF7 / 2.2 kW 6SE7015-0EP50 / 1.5 kW 4500 RPM 17.56 A 1FT6064-6AH7 / 2.3 kW 6SE7018-0EP50 / 3 kW 6000 RPM 14.23 A 1FT6044-4AK7 / 1.9 kW 6SE7015-0EP50 / 1.5 kW In this case, depending on the motor type, the following values are used for i and JG. Rated speed i iopt Nmot max JG 2000 RPM 5 8.48 1500 RPM 0.00084 kgm2 3000 RPM 7 16.12 2100 RPM 0.00024 kgm2 4500 RPM 10 16.33 3000 RPM 0.0002 kgm2 6000 RPM 16 (2stage) 24.08 4800 RPM 0.00018 kgm2 In this case, iopt cannot be used as gearbox ratio due to the dynamic limiting characteristic. The selected drive converters, type of construction Compact Plus were utilized for a 300 % rated current. This is possible, as the current flows for less than 250 ms, and the no-load time is greater than 750 ms. Contrary to example 2 with a fixed gearbox ratio, by optimizing the gearbox ratio, the smallest motor is obtained at the highest rated speed nn=6000 RPM. Siemens AG SIMOVERT MASTERDRIVES - Application Manual 515 4 Information for specific applications 09.99 70 60 M mot, M perm. in Nm 50 40 M perm. (6000 RPM) M perm. (2000 RPM) M perm. (6000 RPM) M perm. (2000 RPM) 30 20 10 0 0 1000 2000 3000 4000 5000 6000 Motor speed in RPM Dynamic limiting characteristics and motor torque curves for motors with nn=2000 RPM (frame size 80) and nn=6000 RPM (frame size 48) 516 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 4.12.6 4 Information for specific applications Example 5, traversing drive with selectable ratio and longer no-load time Drive data Weight to be moved (transported) m = 5500 kg Load wheel diameter D = 0.34 m Specific traversing resistance wF = 0.02 Gearbox ratio i = 16 Mech. efficiency = 0.75 Max. traversing velocity vmax = 2.66 m/s Max. acceleration amax = 0.44 m/s2 Max. traversing travel smax = 42.6 m No-load time after a traversing sequence tP = 10 s In this case, it involves the traversing drive for the high-bay racking vehicle from Section 3.1.4. In this case, a 1PA6 motor is to be used. The following 3 1PA6 motors are selected. Rated speed Max. motor current Motor type / rated output Inverter / rated output 1150 RPM 32.82 A 1PA6107-4HD / 7.2 kW 6SE7022-6TC61 / 11 kW 1750 RPM 30.66 A 1PA6105-4HF / 8 kW 6SE7022-6TC61 / 11 kW 2300 RPM 26.97 A 1PA6103-4HG / 7.5 kW 6SE7022-6TC61 / 11 kW In this case, depending on the motor type, the following values were used for i. Rated speed i iopt Nmot max Nlimit 1150 RPM 7 103 1046 RPM 1313 RPM 1750 RPM 10 103 1494 RPM 1750 RPM 2300 RPM 16 (2stage) 134 2391 RPM 2492 RPM The gearbox ratios were selected, so that the maximum motor speed is as close as possible to speed nlimit. In this case, iopt cannot be used as gearbox ratio, due to the dynamic limiting characteristic and the limit imposed by the maximum permissible motor speed. The gearbox moment of inertia was neglected as a result of the high load moment of inertia. The smallest motor is the 1PA6103-4HG with nn=2300 RPM. This motor also draws the lowest current, as it is the most favorable regarding nlimit and iopt. Motors with nn=400 RPM and nn=2900 RPM cannot be used as the smallest motor with nn=400 RPM is the 1PA6163-4HB, and the smallest motor with nn=2900 RPM, 1PA6184-4HL. Siemens AG SIMOVERT MASTERDRIVES - Application Manual 517 4 Information for specific applications 09.99 In this case, the optimum gearbox ratio is calculated using the motor torque when accelerating M mot max = J mot load max i + ( J load load max + M load ) 1 mech i to J load + iopt = M load load max J mot mech The following is true M load = m g wF load max = amax 0.34 D = 5500 9.81 0.02 = 183.45Nm 2 2 2 2 = 0.44 = 2.59s - 2 D 0.34 D 0.34 2 J load = m ( ) 2 = 5500 ( ) = 158.95kgm 2 2 2 518 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 4 Information for specific applications 120 M mot, M perm. in Nm 100 80 M perm. (2300 RPM) M perm. (1150 RPM) M max (2300 RPM) M max (1150 RPM) 60 40 20 0 0 500 1000 1500 2000 2500 3000 Motor speed in RPM Dynamic limiting characteristics and motor torque characteristics (Mmot max when accelerating) for motors with nn=1150 RPM and nn=2300 RPM Siemens AG SIMOVERT MASTERDRIVES - Application Manual 519 4 Information for specific applications 09.99 4.13 Optimum traversing characteristics regarding the maximum motor torque and RMS torque 4.13.1 Relationships for pure flywheel drives (high-inertia drives) For pure flywheel drives (high-inertia drives), optimum traversing characteristics are to be determined regarding the maximum motor torque and RMS torque. As an example, a rotary table drive will be investigated with the following traversing characteristics. load total load max t tb tv tp t total load load max t - load max Angle tot., positioning time ttot. as well as the no-load interval tp are specified. The traversing characteristic should be symmetrical with tb=tv. 520 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 4 Information for specific applications Then the following is true load max = tot . ttot . - tb load max = load max tb nmot max = for 2 tb ttot . load max 60 i 2 in RPM The maximum motor torque when accelerating or decelerating is then given by M mot max = (( J mot + J G ) i + = (( J mot + J G ) i + J load ) load max i tot . J load ) i tb ttot . - tb2 with JG J load moment of inertia, gearbox i gearbox ratio moment of inertia, load The RMS motor torque and the average motor speed are then given by M rms = 2 M mot max tb 2 ttot . + t p nmot max naverage = 2 = M mot max tb 2 ttot . + t p tb 2 + nmot max (ttot . - 2 tb ) ttot . + t p = tot . 60 i 2 (ttot . + t p ) The maximum motor torque and the RMS torque can be represented as a function of tb, with ttot., tp and tot. specified. The average speed is independent of t b and is given by: M mot max = (( J mot + J G ) i + M rms = M mot max 1 1 J load ) tot . =k 2 i tb ttot . - tb tb ttot . - tb2 2 tb ttot . + t p Siemens AG SIMOVERT MASTERDRIVES - Application Manual 521 4 Information for specific applications 09.99 It will now be investigated for which value of tb, the maximum motor torque is at a minimum. The minimum for Mmot max is given by dM mot max dtb = 0 = -k 1 (tb ttot . - tb2 ) 2 (ttot . - 2 tb ) for tb = ttot . 2 The optimum traversing characteristic for the lowest motor torque when accelerating or decelerating is therefore a triangular characteristic. However, for a triangular characteristic, the highest motor speed is also reached. If the maximum motor torque and the maximum motor speed are referred respectively to the values for the triangular traversing characteristic (tb=ttot./2) then the following is obtained: M mot max = M mot max triangular nmot max nmot max triangular = 1 1 4 tb - ( tb ) 2 ttot . t tot . 1 1 2 1 - tb ttot . 6 M max / M max triangular 5 4 3 2 1 0 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5 tb / t tot. Maximum motor torque as a function of tb/ttot 522 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 4 Information for specific applications 1 n mot max / n mot max triangular 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5 t b / t tot. Max. motor speed as a function of tb/ttot. It will now be investigated, for which value of tb, the RMS motor torque has a minimum. The minimum for Mrms is then given by dM rms =0= dtb dM mot max 2 1 ) ( tb + M mot max ttot . + t p dtb 2 tb 2 t - 2 tb 1 1 (- k tot . tb + k ) 2 2 2 ttot . + t p tb ttot . - tb 2 tb (tb ttot . - tb ) = for tb = ttot . 3 The optimum traversing characteristic for the lowest RMS motor torque is therefore a trapezoidal traversing characteristic with the same accelerating time, decelerating time and constant velocity time components (i.e. same ratios between them). If the RMS motor torque is referred to the values for the triangular traversing characteristic (tb=ttot./2), then the following is obtained: M rms M rms triangular = t 2 1 b ttot . tb - ( tb ) 2 4 ttot . ttot . Siemens AG SIMOVERT MASTERDRIVES - Application Manual 523 4 Information for specific applications 09.99 1.8 M rms / M rms triangluar 1.6 1.4 1.2 1 0.8 0.6 0.4 0.2 0 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5 t b / t tot. RMS motor torque as a function of tb/ttot. 4.13.2 Relationships for flywheel drives (high-inertia drives) with a constant load torque If the same example is used as basis as under Point 4.13.1, then, taking into account a constant load torque, the following changes are obtained. The maximum motor torque is obtained when accelerating, as follows M mot max = (( J mot + J G ) i + = Mb + tot . J load M ) + load 2 i tb ttot . - tb i M load i The RMS motor torque is then given by M rms = 524 (M b + M load 2 M M ) tb + ( load ) 2 (ttot . - 2 tb ) + ( M v - load ) 2 tb i i i ttot . + t p Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 4 Information for specific applications or with Mb=Mv, M rms = M load 2 ) ttot . i ttot . + t p 2 M b2 tb + ( The expression for the average speed does not change. As a result of the constant factor Mload/i at Mmot max, the minimum for the motor torque is again obtained at tb=ttot./2, i. e. for a triangular traversing characteristic. The minimum for the RMS motor torque is then given by Mb = k M rms 1 tb ttot . - tb2 1 2 k2 M = + ( load ) 2 ttot . 2 tb (ttot . - tb ) i ttot . + t p and dM rms 1 =0= dtb ttot . + t p - 2 k2 2 M 2 k2 + ( load ) 2 2 tb (ttot . - tb ) i (ttot . - tb ) 2 - 2 tb (ttot . - tb ) (tb (ttot . - tb2 )) 2 for tb = ttot . 3 Also here, nothing changes. Siemens AG SIMOVERT MASTERDRIVES - Application Manual 525 4 Information for specific applications 4.13.3 09.99 Summary Pure flywheel drives (high-inertia drives) without additional load torques The optimum traversing characteristic for the lowest motor torque when accelerating or when decelerating is a triangular traversing characteristic. However, the highest motor speed is also obtained. If the maximum motor torque is available (e. g. as a result of the dynamic limiting characteristic), then with the triangular traversing characteristic, the shortest positioning time is achieved. The optimum traversing characteristic regarding the lowest RMS motor torque is then obtained for a trapezoidal traversing characteristic with the same sub-division of accelerating time, decelerating time and constant velocity time. Using this traversing characteristic, the lowest motor temperature rise is achieved and a motor speed which is 25% lower with respect to the triangular traversing characteristic; however, the maximum motor torque is 12.5% higher. Flywheel drives (high-inertia drives) with a constant load torque (e. g. friction torque) It can be proven that the triangular traversing characteristic results in the lowest motor torque and the trapezoidal characteristic with t b=ttot./3 results in the minimum RMS value. 526 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 4 Information for specific applications 4.13.4 Example with 1PA6 motor The optimum traversing characteristic for a rotary drive is to be determined. This drive turns tree trunks through 90 degrees within 1 s. This is then followed by a no-load interval of 0.5 s. During this no-load interval, the magnetizing current still continues to flow through the motor. Drive data 2 Load moment of inertia Jload = 1000 kgm Moment of inertia, gearbox JG = 0.00831 kgm Gearbox ratio i = 80 Gearbox efficiency G = 0.95 Mech. efficiency mech = 0.9 Angle tot. = 90 degrees Positioning time ttot. =1s No-load interval tp = 0.5 s 2 As this involves an induction motor, the minimum RMS torque is not investigated, but instead, the minimum RMS current. The condition tb=ttot./3 for the minimum RMS value is also approximately true for the RMS current. The motor torque is given by M mot = (( J mot + J G ) i + J load ) load i (mech G )VZ with VZ = sign( load ) load = load max when accelerating load = - load max when decelerating load = 0 constant velocity motion, no-load interval load max = tot . tb ttot . - tb2 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 527 4 Information for specific applications 09.99 The maximum motor torque when accelerating is given by M mot max = (( J mot + J G ) i + J load ) load max i mech G The following is valid for the motor current (without taking into account saturation): I mot I mot n ( I I M mot 2 1 ) (1 - ( mag ) 2 ) kn2 + ( mag ) 2 2 M mot n I mot n I mot n kn kn = 1 kn = nmot nmot n for nmot nmot n constant flux range for nmot > nmot n field-weakening range The following is valid for the RMS motor current and the average motor speed ( I rms naverage I moti -1 + I moti i ) 2 (ti - ti -1 ) 2 ttot . + t p 60 80 tot . 60 i = = 2 = 800RPM 2 (ttot . + t p ) 2 (1 + 0.5) The two limiting values with tb=ttot./2 (triangular) and tb=ttot./3 (trapezoidal characteristic with t b=tv=tk) is calculated using the PATH configuring program. In this case, both cases are possible for the same motor. The following results are obtained: Triangular trapezoidal (with tb=tv=tk) Mmot max =150.83 Nm Mmot max =169.7 Nm nmot max =2400 RPM nmot max =1800 RPM Irms =53.44 A Irms =49.76 A Imot max =71 A Imot max =76.35 A Pbr max =28.05 kW Pbr max =23.67 kW Pbr average =4.67 kW Pbr average =2.63 kW Braking unit: braking unit: P20=50 kW+ext. brake resistor P20=20 kW+ext. brake resistor 528 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 4 Information for specific applications The same drive converter is also obtained for both of the traversing characteristics. Motor: 1PA6 137-4HG. Pn=29 kW, Mn=120 Nm, Mperm.=240 Nm, nn=2300 RPM, In=56 A, Imag=21 A, Jmot=0.109 kgm2 Drive converter: 6SE7026-0ED61 PU n=30 kW, IU n=59 A, IU max=80.5 A The more favorable case is the trapezoidal characteristic with tb=ttot./3, as the RMS current is lower here (i. e. a lower motor temperature rise), and, in addition, a lower rating braking unit is adequate. For the triangular traversing characteristic, the drive operates in the field-weakening range due to the higher speed. The resulting higher peak current is compensated by the lower maximum motor torque. For the triangular traversing characteristic, a higher braking power is obtained, as the speed increases more than the maximum motor torque decreases. The higher peak current for the trapezoidal traversing characteristic can be explained due to the higher maximum motor torque, and the therefore associated motor saturation effects. 450 400 350 Torque in Nm 300 250 M mot max M perm. 200 150 100 50 0 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5 tb / t tot. Maximum motor torque and dynamic limiting characteristic as a function of tb/ttot. Siemens AG SIMOVERT MASTERDRIVES - Application Manual 529 4 Information for specific applications 09.99 90 80 70 I in A 60 50 I rms In 40 30 20 10 0 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5 tb / t tot. RMS motor current and rated motor current as a function of tb/ttot. 2.5 Angular velocity in 1 / s 2 1.5 1 0.5 0 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 Time in s Angular velocity for the trapezoidal traversing characteristic with tb=ttot./3 530 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 4 Information for specific applications 8 Angular acceleration in 1/s^2 6 4 2 0 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 -2 -4 -6 -8 Time in s Angular acceleration for the trapezoidal traversing characteristic with tb=ttot./3 200 150 Motor torque in Nm 100 50 0 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 -50 -100 -150 Time in s Motor torque for the trapezoidal traversing characteristic with tb=ttot./3 The motor torque when decelerating is lower as a result of the efficiency. Siemens AG SIMOVERT MASTERDRIVES - Application Manual 531 4 Information for specific applications 09.99 40000 30000 Motor power in W 20000 10000 0 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 -10000 -20000 -30000 Time in s Motor output for the trapezoidal traversing characteristic with tb=ttot./3 90.00 80.00 Motor current in A 70.00 60.00 50.00 40.00 30.00 20.00 10.00 0.00 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 Time in s Motor current for the trapezoidal traversing characteristic with tb=ttot./3 532 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 5 5 Index Index A Crawl velocity ............................................. 38, 140 Accelerating force ...............................................11 Cross-cutter drive............................................. 254 Accelerating power .............................................14 Cut length................................................. 254, 272 Accelerating time ..........................................14, 20 Cut lenth, above synchronous.......................... 277 Accelerating time from the line supply ..............395 Cut lenth, sub-synchronous ............................. 275 Accelerating time, shortest .................................17 Cut lenth, synchronous .................................... 276 Accelerating torque .............................................14 Cycle time .......................................................... 36 Accelerating work................................................14 Acceleration ........................................................11 Acceleration due to the force of gravity.................9 Acceleration torque .............................................26 Angle...................................................................11 Angle of rotation..................................................13 Angular acceleration .....................................11, 26 Angular deceleration ...........................................26 Angular velocity...................................................11 Autotransformer ................................................425 B Brake application time.........................................38 Brake control.......................................................35 Brake motor ........................................................38 Brake resistor......................................................36 Braking energy............................................36, 491 Braking time ........................................................20 Braking time, minimum .....................................406 D DC current braking........................................... 463 Dead center...................................... 208, 294, 307 Deadtimes, changing ....................................... 139 Deceleration torque............................................ 26 Deforming work ................................................ 184 Distance ....................................................... 11, 13 Distance traveled ............................................. 144 Drag force .......................................................... 25 Drive wheel ........................................................ 25 Droop ............................................................... 199 E Eccentric press ................................................ 184 Elevator drive ..................................................... 83 Elevator height ................................................... 83 Emergency off .................................................. 430 Energy.................................................................. 9 Energy equation ............................................... 492 Breakaway torque .............................................465 F C Cable drum .........................................................29 Capacitor battery...............................................284 Centrifugal drive................................................261 Connecting bar force.................................186, 213 Conservation of energy law...............................191 Constant power.................................................192 Contact force.....................................................313 Continuous braking power ..................................36 Crank angle.......................................................185 Crank drive........................................................208 Siemens AG SIMOVERT MASTERDRIVES - Application Manual Feed force .......................................................... 18 Feed velocity ...................................................... 18 Field-weakening range............................. 192, 392 Flywheel ........................................................... 184 Force .................................................................... 9 Force-ventilated ................................................. 33 Four-jointed system.......................................... 306 Frame size, minimum....................................... 499 Friction angle, spindle ........................................ 18 Friction coefficient .............................................. 27 Full-load steady-state output.............................. 29 533 5 Index 09.99 G M Gantry crane....................................................... 77 Mass..................................................................... 9 Gearbox ............................................................. 16 Master/slave drive ............................................ 199 Gearbox efficiency ............................................. 16 Mechanical brake ............................................. 491 Gearbox ratio ..................................................... 16 Mechanical efficiency ......................................... 26 Gearbox ratio, optimum ................................... 503 Mechanically-coupled drives ............................ 197 H Harmonics and disturbances fed back into the line supply ........................................ 483 Mesh welding machine..................................... 323 Moment of inertia.......................................... 9, 161 Moment of inertia, roll....................................... 106 High-bay racking vehicle .................................... 60 Moment of inertia, spindle .................................. 18 Hoisting drive................................................ 25, 34 Motor utilization ................................................ 499 hoisting force...................................................... 13 Multi-motor drives............................................. 474 Hoisting force ..................................................... 29 Hoisting unit........................................................ 77 N No-load interval time .......................................... 33 Hoisting work...................................................... 13 Holding brake ............................................. 35, 491 O Oscillation frequency, mechanical.................... 174 I Inclination angle ................................................. 92 Overhead wire height ....................................... 357 Overlap angle ................................................... 274 Incline plane ....................................................... 92 Intermittent duty.................................................. 32 Isolating transformer ........................................ 427 P Pantograph....................................................... 357 Partial load range ............................................. 418 K Kinetic buffering ....................................... 398, 474 Kinetic energy..................................................... 11 Knife roll ........................................................... 254 Permissible acceleration .................................... 28 Pitch angle, spindle ............................................ 18 Pivot drive......................................................... 227 Pivot motion...................................................... 227 L Position change, periodic ................................. 151 Leakage inductance ......................................... 418 Position controller ............................................. 141 Line failure........................................................ 398 Position difference.................................... 141, 143 Linear motion ..................................................... 11 Positioning drives ............................................. 139 Load distribution ............................................... 197 Positioning error ............................................... 140 Load duty cycle ............................................ 32, 36 Positioning time ................................................ 145 Load equalization control ................................. 200 Positioning, closed-loop control........................ 139 Load mass.......................................................... 29 Positioning, closed-loop control........................ 149 Load sinking ....................................................... 35 Positioning, fastest ........................................... 144 Load torque ................................................ 16, 186 Positioning, open-loop control .................. 139, 147 Load torque characteristic................................ 361 Power ................................................................... 9 Load torque oscillation ..................................... 153 Power adaption................................................. 502 Load torque, square-law .................................. 387 Power-on duration .............................................. 32 Press drive ....................................................... 184 Press force ....................................................... 184 534 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 09.99 5 Index R Static friction....................................................... 27 Ram distance ....................................................186 Supplementary moment of inertia .................... 152 Ram time (surge time) ......................................187 Surge loading ................................................... 151 Re-accelerating time .........................................398 Switch-on duration ............................................. 32 Reciprocating compressor, single-cylinder .......161 Synchronous speed ............................................. 9 Reduction factor kf ..............................................10 System losses .................................................. 453 Reel...................................................................103 Regenerative operation...............................36, 483 Resistance force .................................................25 Risk evaluation..................................................430 RMS motor current ...........................................195 RMS torque...................................................10, 32 Roll radius ...........................................................13 Rotary table drive ..............................................221 Rotational............................................................11 Rotational energy................................................11 Rounding-off .....................................................442 T Tangential force ....................................... 186, 213 Three-cylinder pump ........................................ 167 Time constant, mechanical .............................. 152 Torque.................................................................. 9 Torque compensation ...................................... 174 Torque limit ...................................................... 174 Traction drive ............................................... 25, 34 Transformer at the drive converter output........ 425 Traversing curves, optimum............................. 520 Trolley .............................................................. 171 Rraction cycle .....................................................28 U S Saw drive ..........................................................293 Unwind stand with closed-loop tension control........................................................... 110 Self locking..........................................................19 Unwinding ........................................................ 105 Self-ventilated .....................................................33 Shaft output...........................................................9 V Velocity............................................................... 11 Sinusoidal filter..................................................425 Slip ........................................................................9 W Slip frequency .......................................................9 Web velocity..................................... 103, 254, 272 Specific traction resistance .................................25 Wind force.......................................................... 26 Speed (RPM) ......................................................13 Wind pressure.................................................... 26 Spindle diameter .................................................18 Wind surface ...................................................... 26 Spindle drive ...............................................18, 238 Winder control.................................................. 117 Spindle efficiency ................................................18 Winder drive..................................................... 103 Spindle length .....................................................18 Winder with closed-loop tension control .. 109, 117 Spindle pitch .......................................................18 Winding ............................................................ 105 Spindle, horizontal.............................................246 Winding power ................................................. 104 Spindle, vertical.................................................238 Winding torque................................................. 104 Stall torque....................................................9, 192 Work..................................................................... 9 Siemens AG SIMOVERT MASTERDRIVES - Application Manual 535 6 Literature 6 /1/ 09.99 Literature Moderne Stromrichtertechnik Peter F. Brosch Vogel Fachbuch Kamprath-Reihe ISBN 3-8023-0241-9 /2/ Grundlagen der elektrischen Antriebstechnik mit Berechnungsbeispielen J. Vogel Huthig Verlag, 1991 ISBN 3-7785-2103-9 /3/ AC-Servo-Antriebstechnik Lehmann, R. Franzis-Verlag, 1990 ISBN 3-7723-6212-5 /4/ Elektrische Vorschubantriebe fur Werkzeugmaschinen H. Gro Siemens AG ISBN 3-8009-1338-0 /5/ Das 1x1 der Antriebsauslegung W. Garbrecht, J. Schafer VDE Verlag, 1994 ISBN 3-8007-2005-1 /6/ Lenze Formelsammlung Carsten Frager Lenze GmbH&Co KG Aerzen /7/ Technische Formeln fur die Praxis Walter (u.a.) Buch-und Zeit-Verlagsgesellschaft mbH Koln, 1992 ISBN 3-8166-0192-8 536 Siemens AG SIMOVERT MASTERDRIVES - Application Manual