减速电机选型.pdfVIP

NORD - Information Calculation Methods and Examples Getriebebau NORD, Schlicht + Kchenmeister GmbH & Co. Rudolf-Diesel-Str. 1, D- 22941 Bargteheide Telefon: 04532 / 4010, Telefax: 04532 / 401 253 NORD - Information Physical Formulae Linear motionRotating motion Distances = v tAngular = t Velocity (speed)v = s t Angular velocity = = 2 n Accellerationa = v t Angular accelleration = W t ForceF = m aTorqueM = J r = F r PowerP = F vPowerP = M WorkW = F s = P tWorkW = M = P t Kinetic energyWkin = 1 2 m v 2 Rotating energyWrot = 1 2 J 2 Formulae of drive engineering Rolling resistance, -forceFR = m g 2 D (L d 2 + f) + c : L, f, c, s. tables_ _ _1 _, _ _2 _, _ _3 _. _ orFR = We mWe for wheel / rail steel s. diagramme 1, 2. Sliding resitance, -forceFG = m g s. table 4 Static friction-forceFH = m g OO s. table 4 WindloadF = A PW Moment of inertia with refernceJred = 91,2 m v nM 2Translation to the motor shaftorJred = J n nM 2Rotation Speedn = v 60 D TorqueM = P 9550 n Friction powerPR = F v 1000 Translation orPR = M n 9550 Rotation Acceleration powerPB = m a v 1000 Translation orPB = J n2 91,2 1000 tB Rotation 2 NORD - Information a aB av A c d dO d1 d2 D f fB fZ F FG FH FQ FQvorh FQzul FR FW g i iV J JBre JM Jred JZ LN m maf mG mL mO Formulae symbols and unities Acceleration Acceleration (start up) Deceleration (braking) Area (wind) Additional factor for secondary friction Diameter (bearing spigot diameter) Pinion or sprocket diameter Pinion diameter Chain sprocket diameter Diameter of the travelling wheel or cable drum or of the sprocket Lever arm of rolling friction Service factor Additional factor for overhung load Force, rolling resistance Sliding friction Static friction Overhung load Existing overhung load Permissible overhung load Rolling resistance Wind load Gravity (constant: 9,81) Reduction Additional reduction (gear, chain, belt .) Moment of inertia Moment of inertia of the brake Moment of inertia of the motor Moment of inertia with reference to the motorshaft Moment of inertia of the z-fan Brake service life until readjustment Weight (mass) Inertia mass acceleration factor Mass of counter weight Mass with full load Mass without load m/s2 m/s2 m/s2 m2 - m m m m m m - - N N N N N N N N m/s2 - - kgm2 kgm2 kgm2 kgm2 kgm2 h kg - kg kg kg 4 NORD - Information M MB MH ML MN M1 M2 M2max n nM nN n1 n2 PW P Pab Pzu PB PHub PN PR r s sB sV t tB tV v W We Wkin Wrot Wzul WB x Torque Braking torque Run up torque Torque with full load (with reference to the motor shaft) Rated torque Input torque Output torque Maximum permissible output torqe Speed Motor speed Rated speed Input speed Output speed Wind pressure Power Required power Supplied power Acceleration power Lifting power Rated power Fricition power Radius Distance Start up distance Braking distance Time Start up time Braking time Velocity (speed) Work Standard rolling friction Kinetic energy Rotating energy Braking work until readjustment Braking work Number of drives Nm Nm Nm Nm Nm Nm Nm Nm 1/min 1/min 1/min 1/min 1/min N/m2 kW kW kW kW kW kW kW m m m m s s s m/s J N/t J J J J - 5 NORD - Information z zzul zO z1 z2 G G L O Starting frequency Permissible starting frequency Starting frequency with no load Number of gear teeth pinion Number of gear teeth gear wheel Angular acceleration Efficiency Efficiency of gear unit Reverse operating efficiency Coefficient of friction Coefficient of friction for bearings Coefficient of friction (static) Angular Angular velocity s/h s/h s/h - - 1/s2 - - - - - - 1/s 6 NORD - Information table 1: coefficient of friction for bearings L table 2: lever arm of rolling friction f table 3: rim friction on the wheels c table 4: static friction and sliding friction table 5: efficiency table 6: additional factor inderming overhung loads fz Friction bearingSliding bearings L0,0050,1 7 f steel / steel0,0005 m wood / steel0,0012 m polymer / steel0,002 m hardrubber / steel0,0077 m hardrubber / concrete0,01 - 0,02 m rubber / concrete0,015 - 0,035 m Friction of anti friction bearings Friction of sleeve bearings Friction of guide rollers c0,0030,0050,002 Static friction OSliding friction drygreaseddrygreased steel / steel0,11 - 0,400,100,10 - 0,300,01 - 0,10 steel / last iron0,18 - 0,250,100,16 - 0,250,05 - 0,10 steel / wood0,50 - 0,700,100,20 - 0,500,02 - 0,10 steel / polymer0,20 - 0,500,10 - 0,35 steel / rubber0,40 - 0,50 wood / wood0,40 - 0,800,160,20 - 0,500,04 - 0,16 fZ helical gears1,1z = 17 teeth chain sprockets1,4z = 13 teeth chain sprockets1,2z = 20 teeth pulleys1,7by tensioning influence pulleys2,5by tensioning influence chain0,90 - 0,96per complete wrap of the rope around the drum wire ropes0,90 - 0,95per complete wrap flat polymer belts0,93 - 0,98per complete wrap of the rope depending on the material V-belts0,85 - 0,95per complete wrap rubber belts0,80 - 0,85per complete wrap polymer belts0,80 - 0,85per complete wrap helical inline gear0,95 - 0,98oil lubricated depending on the number of the stages worm gear0,30 - 0,93oil lubricated depending on the number of starts of the worm NORD - Information Additional for rim friction : 30 N / t Additional for rim friction : 50 N / t Diagram 1: Standard rolling friction We of anti friction Diagram 2: Standard rolling friction We of anti friction 8 NORD - Information Example I.1: Drive arrangement for crane Mass without load of the crane mO13800 kg Mass without load of the mk1800 kg Load mL15000 kg Velocity v0,17 / 0,66 m/s = 10/40 m/min Diameter of the travelling wheel D0,4 m Number of drives x2 Additional reduction iv4,24 Mounting positionB 3 Switching frequencies z60 s/h Efficiency 0,85 9 NORD - Information Motor arrangements Standard rolling friction We We = WO + 30 N/tW0 = 36 N/t s. diagram We = 36 N/t + 30 N/t = 66 N/t 30 N/t additional for rim friction Power P (at maximum velocity) P = We m v 1000 PO = 66 N t (13,8 t + 1,8 t) 0,66 ms 1000 0,85 = 0,80 kW (without load) PL = 66 N t (13,8 t + 1,8 t + 15,0 t) 0,66 ms 1000 0,85 = 1,57 kW (with load) Pmax = PL 2 mO + 2 (mK + mL ) mO + mK + mL = 1,57 kW 2 13,8 t + 2 (1,8 t + 15 t) 13,8 t + 1,8 t + 15 t Pmax = 1,22 kW (one-sided trolley) Motor data Type100 L/80-20 WU Bre16 Z (2 pieces) Rated output power PN0,55 / 2,2 kW Rated speed nN670 / 2740 1/min Rated torque MN7,8 / 7,7 Nm Permissible no-load starting frequency zo4000 / 1400 s/h Motor moment of inertia JM0,0060 kgm2 Additonal moment of inertia Jz0,0113 kgm2 Brake moment of inertia JBre0,0001 kgm2 Braking torque MB (brake 16 adjusted to 8 Nm )8 Nm 10 NORD - Information Gear arrangment Wheel speed nL nL = v 60 D nL = 0,66 ms 60 0,4 m = 32 1min Gear output speed n2 n2 = nL iv n2 = 32 1min 4,24 = 136 1min Acceleration factor of mass maf maf = Jred JM + Jz + JBre maf = 0,0810 kgm2 0,0060 kgm2 + 0,0113 kgm2 + 0,0001 kgm2 = 4,7 Starting freqeuency per hour: 180 (60 times acceleration, switching, deceleration) Type of load C, fB = 1,6 Output torque Ma Ma = PN 9550 n2 fB Ma = 2,2 kW 9550 136 1min 1,6 = 247 Nm For service factor fB = 1,6 the output torque of the gear is 247 Nm. Reduction i i = nN n2 i = 2740 1min 136 1min = 20 Complete type:SK 22-100 L/80-20 WU Bre 16 Z PN = 0,55 / 2,2 kW i = 20,03 n2 = 33 / 137 1/min Mounting position B 3 Shaft 30 x 60 mm Brake 16 Nm adjusted to 8 Nm Special provision:special rotor high inertia fan 11 NORD - Information Example I.2: Drive arrangement for a trolley Mass without load mk1800 kg Load mL15000 kg Velocity v0,08 / 0,33 m/s = 5/20 m/min Wheel diameter D0,3 m Number of drives x1 Additional reduction iv4 Mounting positionB 5 Switching frequency z60 s/h Efficiency n 0,85 Pairing of materialsteel / steel Guidingrim friction Type of bearings (4 wheels)antifriction bearings 12 NORD - Information Drive resistance: FR = m g 2 D ( L d 2 + f ) + c Fro = 1800 kg 9,81 m s2 2 0,3 m (0,005 0,06 2 + 0,0005 m ) + 0,003m Fro = 129,5 N ( without load) FRL = 16800 kg 9,81 m s2 2 0,3 m (0,005 0,06 2 m + 0,0005 m ) + 0,003 m FRL = 1208,6 N (with load) Power P (calculation for 2-poles gearmotors) P = F V 1000 Po = 129,5 N 0,33 m 1000 0,85 s = 0,05 kW PL = 1208,6 N 0,33 1000 0,85 = 0,47 kW 13 NORD - Information Motor arrangement Motor data Type100 L/8-2 WU Bre10 Z Rated output power PN0,4 / 1,6 kW Rated speed nN670 / 2740 1/min Rated torque MN5,7 / 5,6 Nm Hochlaufmoment MH9,2 / 8,6 Nm No-load switching frequency zo4200 / 1500 s/h Motor moment of inertia JM0,0045 kgm2 Moment of high inertia fan Jz0,0113 kgm2 Brake moment of inertia JBre0,0001 kgm2 Braking torque MB (brake 10 adjusted on 6 Nm)6 Nm Load torque M M = P 9550 x nN MO = 0,05 KW 9550 2740 1min = 0,2 Nm (without load) ML = 0,47 kW 9550 2740 1min = 1,6 Nm (with load) Reduced moment of inertia Jred Jred = 1 x 91,2 m v nN 2 JredO = 91,2 1800 kg 0,33 ms 2740 1min 2 = 0,0024 kgm2 JredL = 91,2 16800 kg 0,33 ms 2740 1min 2 = 0,0222 kgm2 Acceleration aB aB = 9,55 v (MH ML) (Jred + JM + JBre + JZ) nN aB = 9,55 0,33 ms (8,6 Nm 0,2 Nm) (0,0024 kgm2 0,85 + 0,0045 kgm2 + 0,0001 kgm2) 2740 1min = 0,52 ms2 (without load) aB = 9,55 0,33 ms (8,6 Nm 1,6 Nm) (0,0222 kgm2 0,85 + 0,0045 kgm2 + 0,0001 kgm2) 2740 1min = 0,19 ms2 (with load) 14 NORD - Information Decceleration aV av = 9,55 v (MB + ML 2) (Jred + JM + JBre + JZ) nN aVO = 9,55 0,08 ms (6 Nm + 0,2 Nm 0,852) (0,0024 kgm2 0,85 + 0,0045 kgm2 + 0,0001 kgm2 + 0,0113 kgm2) 670 1min = 0,39 ms2 (without load) aVL = 9,55 0,08 ms (6 Nm + 1,6 Nm 0,852) (0,0222 kgm2 0,85 + 0,0045 kgm2 + 0,0001 kgm2 + 0,0113 kgm2) 670 1min = 0,24 ms2 (with load) Permissible switching frequency zzul zzul = 1 ML MH 1 + (Jred + JZ + JBre) JM z0 zzul = 1 1,6 Nm 8,6 Nm 1 +(0,0222 kgm2 + 0,0113 kgm2 + 0,0001 kgm2) 0,0045 kgm2 1500 sh = 142 sh The perm. switching frequency is calculated for the acceptance: starting 2-pole with load (every time) is not correct because of delay for switching and running on the 8-pole. 15 NORD - Information Gear arrangements Wheel speedl nL nL = v 60 D nL = 0,33 ms 60 0,3 m = 21 1min Gear unit output speed n2 n2 = nL iv n2 = 21 1min 4 = 84 1min Mass acceleration factor maf maf = Jred JM + Jz + JBre maf = 0,022 kgm2 0,0045 kgm2 + 0,0113 kgm2 + 0,0001 kgm2 = 1,4 Circuit m per hour: 180 (each 60 accelerations, switching, decelerations) type of load B fB 1,3 Output torque Ma Ma = PN 9550 n2 fB Ma = 1,6 kW 9550 84 1min 1,3 = 236 Nm For service factor fB = 1,3 the output torque of the gear is 236 Nm. Reduction i i = nN n2 i = 2740 1min 84 1min = 33 16 NORD - Information Complete type:SK 22 F - 100 L/8-2 WU Bre 10 Z PN = 0,4 / 0,16 kW i = 34,69 n2 = 19/79 1/min Mounting position B 5 Shaft 30 x 60 mm Flange 160 mm oder 200 mm Brake 10 Nm adjusted on 6 Nm Special provision:special rotor (WU-silumin rotor) high inertia fan 17 NORD - Information Example II.1: Drive unit for vertical motion Mass without load mO50 kg Load mL200 kg midle drum diameter Dm0,208 m Max. lifting speed v0,24 m/s = 14,4 m/min Operation cycle8 h/Tag, 40 % ED Starting frequency z360 Hubbewegungen/h Efficiency 0,8 Positioning accuracy 1 mm 18 NORD - Information Motor arrangement Power P P = m g v 1000 PL = (50 kg + 200 kg) 9,81 ms2 0,24 ms 1000 0,8 = 0,74 kW To get the required accuracy of 1 mm we have to choose a polechanging motor. Motor data Typ80 L/4-2 Bre8 Rated output power PN0,60 / 0,75 kW Rated speed nN1400 / 2830 1/min Synchronous speed nsyn1500 / 3000 1/min Rated torque MN4,1 / 2,5 Nm Run-up torque MH7,4 / 5,7 Nm No-load switching frequency zo2500 / 1800 s/h Motor moment of inertia JM0,00165 kgm2 Brake moment of inertia JBre0,00007 kgm2 max. braking work until readjustment Wzul.7 * 107 J Brake reaktion time t20,015 s (DC-connection) Braking torque MB8 Nm Load torque M M = P 9550 nN ML = 0,74 kW 9550 2830 1min = 2,5 Nm Switching torque MU MU = 2 * MH4 MU = 2 * 7,4 Nm = 14,8 Nm reduced moment of inertia Jred Jred = 91,2 m v nN 2 Jred = 91,2 (50 kg 200 kg) 0,24 ms 2830 1min 2 = 0,00016 kgm2 19 NORD - Information z0 = 2320 / 1620 s/h = max. perm. switching frequency with no load For this application A 4-2 polemotor (Dahlander-connection) is used. Therefor the half of the zo is criteria. Permissible switching frequency zzul up motion: zzul = 1 ML MH 1 + (Jred + JBre) JM zO 2 zzul = 1 2,5 Nm 5,7 Nm 1 + (0,00016 kgm2 + 0,00007 kgm2) 0,00165 kgm2 1620 sh 2 = 399 sh (2 poles) down motion: zzul = 1 ML MU 1 + (Jred + JBre) JM zO 2 zzul = 1 2,5 Nm 14,8 Nm 1 + (0,00016 kgm2 + 0,00007 kgm2) 0,00165 kgm2 2320 sh 2 = 846 sh (4 poles) The mechanical braking depends on the positioning speed. The max. braking distance depends on down motion. Deceleration av av = 9,55 v nN4 nN2 (MB ML 2) (Jred + JM + JBre) nN4 av = 9,55 0,24 ms 1400 1min 2830 1min (8 Nm 2,5 Nm 0,82) (0,00016 kgm2 0,8 + 0,00165 kgm2 + 0,00007 kgm2) 1400 1min = 2,80 ms2 In case of calculation the deceleration time we have to use the increased speed for the down motion. The cause is the delay for switching and the over-synchronous speed. Load speed nL nL = nsyn ML MN (nsyn nN)+: down motion, -: up motion down motion: nL = nsyn + ML 2 MN (nsyn nN) nL = 1500 1min + 2,5 Nm 0,8 2 4,1 Nm (1500 1min 1400 1min) = 1539 1min 20 NORD - Information Increased speed during braking time n n = 9,55 ML t2 Jred + JM + JBre +: down motion, -: up motion down motion: n = 9,55 ML 2 t2 Jred + JM + JBre n = 9,55 2,5 Nm 0,82 0,015 s 0,00016 kgm2 0,8 + 0,00165 kgm2 + 0,00007 kgm2 = 124 1min Deceleration time tv (Braking time) tv = v (nL + n) nN2 a tv = 0,24 ms (1539 1min + 124 1min) 2830 1min 2,80 ms2 = 0,05 s Deceleration distance sv (Braking distance) sv = v nL + n nN2 2 2 a sv = 0,24 ms 1539 1min + 124 1min 2830 1min 2 2 2,80 ms2 = 0,004 m Positioning accuracy The positioning accuracy is about 25 % from the deceleration distance sv. Positioning accuracy = 25 % * sv = 0,25 * 0,004 m = 0,001 m Braking work WB WB = (Jred + JM + JBre) n 2N4 182,5 MB MB ML WB = (0,00016 kgm 2 0,8 + 0,00165 kgm2 + 0,00007 kgm2) (1400 1min)2 182,5 8 Nm 8 Nm 0 = 20 J Because of the same number of up- and down-motion the load torque = 0 Nm. Brake service life until readjustment LN LN = Wzul WB z LN = 7 107J 20 J 360 1h = 9720 h 21 NORD - Information Gear arrangements Gear output speed n2 n2 = v 60 Dm n2 = 0,24 ms 60 0,208 m = 22 1min Mass acceleration factor maf maf = Jred JM + JBre maf = 0,00016 kgm2 0,00165 kgm2 + 0,00007 kgm2 = 0,09 Switching per hour: 1080 ( each 360 accelerations, change-over, decelerations) kind of load A, fB = 1,2 Output torque Ma Ma = PN 9550 n2 fB Ma = 0,75 kW 9550 22 1min 1,2 = 391 Nm Reduction i i = nN n2 i = 2830 1min 22 1min = 129 Complete type:SK 2382 A - 80 L/4-2 Bre8 PN = 0,60 / 0,75 kW i = 131,86 n2 = 11 / 21 1/min Mounting position H 1 Hollow shaft 35 mm Brake 8 Nm Insolating material class F 22 NORD - Information Example III.1: Turntable drive for processing table Determine the size of a cd-geared motor for a tuntable with 3 work stations ( = 120) Table weight without loadmO500 kgPositioning accuracy= 1 mm Table diameter D2 mSprocket reductioniv3,76 Positions of load 120 Duty factorED60 % Spaced at radiusR1 mPulse number360 Takte/h Ball bearing ring diameter d2 mTime of run16 h/Tag Cycle time for 120 turntges6 sEfficiency 0,8 Load mL (3 x 750 kg)mL2250 kgMounting positionV 6 23 NORD - Information Distance s ( at a rotation of 120 ) s = D 3 = 2 m 3 = 2,094 m Acceleration time tB or Deceleration time tV tB = tV = 1 s (acceptance data) Table speed nT nT = sges 60 D ( t ( tB + tV ) 2 ) = 2,094 m 60 2 m ( 6 s ( 1 s + 1 s ) 2 ) = 4 1 min Table circumferential velocity v (Ball bearing ring) v = d nT 60 = 2 m 4 1min 60 = 0,42 ms Moment of inertia J J = 1 8 mO D 2 + 1 4 mL d 2 = 1 8 500 kg (2 m) 2 + 1 4 2250 kg (2 m) 2 = 2500 kgm2 Friction power PR (static) PR = (mO + mL) g L v 1000 = (500 kg + 2250 kg) 9,81 ms2 0,005 0,42 ms 1000 0,8 = 0,07 kW with L = 0,005 for friction bearing Acceleration power PB (dynamic) PB = J nT2 91,2 1000 tB = 2500 kgm 2 (4 1min)2 91,2 1000 1 s 0,8 = 0,55 kW Power P P = PR + PB (friction + acceleration) P = 0,07 kW + 0,55 kW = 0,62 kW 24 NORD - Information Motor data Type90 S/8-2 Bre 10 Rated power PN0,25 / 1,1 kW Rated speed nN700 / 2810 1/min Rated torque MN3,4 / 3,7 Nm Hochlaufmoment MH4,0 / 5,7 Nm No-load switching frequency zo9000 / 1500 s/h Motor moment of inertia JM0,00235 kgm2 Brake moment of inerta JBre0,00007 kgm2 Braking torque MB (adjusted at 8 Nm)8 Nm Load torque ML M = PR 9550 nN = 0,07 kW 9550 2810 1 min = 0,2 Nm Reduced moment of inertia Jred Jred = J nT nN 2 = 2500 kgm2 4 1min 2810 1min 2 = 0,00507 kgm2 Permissible switching frequence zzul zzul = 1 ML MH 1 + (Jred + JBre) JM zO = 1 0,2 Nm 5,7 Nm 1 + (0,00507 kgm2 + 0,00007 kgm2) 0,00235 kgm2 1500 sh = 453