减速机详细的选型计算及练习

03 04 201G tz Product Support Drive selection 目录目录 Contents 1练习简介练习简介 Brief description of the exercises 1 2实用工具实用工具 Aids 2 3练习练习 Exercises 3 3 1结构设计形式为 M 的减速电机 Geared motor design M 3 3 2结构设计动工为 N 的减速电机 Geared motor design N 4 3 3制动单元练习 1 Braking unit 1 5 3 4制动单元练习 2 Braking unit 2 6 3 5传动轴 Spindle 7 4练习答案练习答案 Solutions 8 4 1结构设计形式为 M 的减速电机 Geared motor design M 8 4 2结构设计形式为 N 的减速电机 Geared motor design N 10 4 3制动单元练习 1 Braking unit 1 12 4 4制动单元练习 2 Braking unit 2 14 4 5 传动轴 Spindle 15 1 练习练习简简介介 Brief description of the exercises 主题 Topic 内容 Contents 难度 Grade 传动轴驱动 Spindle drive往复传动轴 Cyclic spindle drive 需选择 电机 变频器 制动单元 searched Motor inverter braking unit 高 High 制动单元 Braking unit 加速传动 Acceleration drive 一般 Normal 制动单元 Braking unit 提升 电阻混联 Hoist resistor network 高 High 直流母线 DC bus 直流母线运行实例 Examples of DC bus operation 一般 Normal 减速电机 Geared motor 设计形式为 M 的 Lenze 减 速电机的选型 Catalogue selection of a Lenze geared motor in M design 一般 Normal 减速电机 Geared motor 设计形式为 N 的 Lenze 减 速电机的选型 Catalogue selection of a geared motor in N design 一般 Normal 横切机 Cross cutter 卷曲设备 Winder 提升设备 Lift drive 传输设备 Travelling drive 03 04 202G tz Product Support Drive selection 2实用工具实用工具 Aids 计算器 Pocket calculator Lenze 选型手册 Lenze catalogues Lenze 公式集 Lenze formula collection 03 04 203G tz Product Support Drive selection 3练习练习 Exercises 3 1 设计形式为设计形式为 M 的的 Lenze 减速电机的选型减速电机的选型 Geared motor design M 减速电机按 S2 方式进行传动 运行时间 10min 此时 可采用常规运行方式 A geared motor is to drive a load in S2 operation operating time 10 min In this case a regular operation is given 具体数据具体数据 Detailed data 转矩 Process torque M2 580 Nm 速度 Process speed n2 100 rev min 主电压 Mains voltage VMains 400 V 主电源频率 Mains frequency fMains 50 Hz 运行时间 Operating time day 8 h 所需部件所需部件 Searched components Lenze 异步电机 Lenze asynchronous motor GST 减速器 Gearbox GST 03 04 204G tz Product Support Drive selection 3 2 设计形式为设计形式为 N 的的 Lenze 减速电机的选型减速电机的选型 Geared motor design N 减速电机按 S2 方式进行传动 运行时间 10min 此时 可采用常规运行方式 A geared motor is to drive a load in S2 operation operating time 10 min In this case a regular operation is given 具体数据具体数据 Detailed data 转矩 Process torque M2 580 Nm 速度 Process speed n2 100 rev min 主电压 Mains voltage VMains 400 V 主电源频率 Mains frequency fMains 50 Hz 运行时间 Operating time day 8 h 所需部件所需部件 Searched components Lenze 异步电机 Lenze asynchronous motor GST 减速器 Gearbox GST 注 N 型减速器可用于 IEC 连接 作为规则连接 该型电机应为外置式 为便于计算 可选用 Lenze 电机 Note Type N is designed for motors with an IEC connection As a rule these are external motors To make calculating easier Lenze motors can be used for this calculation 03 04 205G tz Product Support Drive selection 3 3 制动单元制动单元 1 Braking unit 1 Process 利用伺服控制对圆柱型固体进行加速及制动的驱动特性如上图所示 A solid cylinder is accelerated and braked by a servo drive as shown in the above characteristic 具体数据具体数据 Detailed data 圆柱体质量 Mass of the cylinder m 2 kg 圆柱体半径 Radius of the cylinder r 0 25 m 摩擦转矩 riction torque MFriction 3 Nm 最大速度 Max speed n 2500 rpm 加速时间 Acceleration time t1 2 s 延迟时间 Delay time t3 1 s 静止周期 Rest period t4 1 s 循环周期 Cycle time T 7 s 电机功效 Efficiency of the motor Motor 0 8 电机转动惯量 Moment of inertia of the motor JMotor 10 kgcm2 变频器功耗 Power loss of the inverter PV 260 W 需选择需选择 Searched components 制动单元 Braking unit resistor 转矩及功率曲线 Torque and power profile t1t2t3t4 T n t 03 04 206G tz Product Support Drive selection 3 4 制动单元制动单元 2 Braking unit 2 电机 Motor 两台 37kW 电机 忽略功效 安全预留 2 motors with 37 kW efficiency neglected safety reserve 控制器 Controller 两台 EVF9200ES 忽略功耗 安全预留 2 pieces of the EVF 9330 ES power loss neglected safety reserve 质量 Mass m 130 000 kg 高度 Height h 55 m 速度 Speed v 3 m min 接触倾角 No contact bevel angle 0 应用范围 Application Hoist without counter weight 需选择需选择 Searched components 制动单元 制动电阻 Braking unit resistor 03 04 207G tz Product Support Drive selection 3 5 传动轴传动轴 Spindle 应用应用 Application 传动轴用于延固定轨迹传送一刚体 此时 传动往复路径是一致的 刚体安装在导轨 上 The spindle is to move a mass of steel according to a specified profile In this case the return trip is the same The mass is mounted on rails 具体数据具体数据 Detailed data 材料质量 Material mass 1 5 t 前进距离 Forward feed distance 240 mm 传动轴材料 Spindle material 钢 steel 传动轴倾度 Spindle pitch 10 mm 传动轴摩擦直径 Spindle friction diameter 28 mm 传动轴类型 Spindle type 球轴承 ball bearing spindle 传动轴长度 Spindle length 900 mm 传输速度 Traversing speed 12 m min 加速时间 Acceleration time 0 3 s to 0 5 s 延迟时间 Delay time 0 3 s to 0 5 s 静止周期 Rest period 0 1 s 与导轨之前的摩擦系数 Friction coefficient of the rails b 0 02 需选择需选择 Searched components 异步电机 不带减速器 Asynchronous motor without gearbox 变频器 矢量型 Frequency inverter vector 制动斩波器 制动电阻 Brake chopper resistor0 03 04 208G tz Product Support Drive selection 4练习答案练习答案 Solutions 4 1 设计形式为设计形式为 M 的的 Lenze 减速电机的选型减速电机的选型 Geared motor design M 求传输功率 Calculation of the process power 4 1 W nM P75 6073 60 2 22 2 求 kS2 1 4 且 Gearbox initial 0 95 时所需的电机功率 Calculation of the required motor power with kS2 1 4 and Gearbox initial 0 95 4 2 W k P P initialgearboxS req 73 4566 2 2 1 根据主电源数据选择电机电压及频率 Motor voltage and motor frequency correspond to the mains data 可选择 4 极异步电机 112 32 A 4 pole asynchronous motor is selected 112 32 电机型号 Motor size PNnrIrIA IrV frcos MrMstallMA J kW min 1 A A V Y Hz Nm Nm NM 10 3 kgm2 112 325 5144012 58 0 400 2 500 788936 5138 7105 922 8 供电电压 400V 连接方式 角接 Delta interconnection with 400 V 求减速器速比 Calculation of the setpoint gearbox ratio 4 3 4 14 2 n n i N soll 负载等级为 I Load class I is defined 由于在 S2 方式下运行 10 分钟 故每小时开关次数很少 The number of operations per hour is very small because of the S2 operation of 10 minutes 运行因子最大为 0 9 This leads to an operation factor k of max 0 9 根据 G motion const 手册中 14 286 c 1 3 查出 iactual Selection of iactual in the G Motion const catalogue of14 286 c 1 3 此时 In this case c k GST07 2M 若所需传递的转矩传至电机侧 则结果为 Gearbox 0 97 If the requested process torque is transformed to the motor side the result is as follows Gearbox 0 97 4 4 Nm i M M Getriebeist 85 41 2 2 可根据电机的运行值求出 C C could be recalculated based on the operating point of the motor 4 5 13 1 2 M M cc N new 03 04 209G tz Product Support Drive selection 为校核启动转矩 必须将 M2 作为MA 为获得充足的加速裕量 必须确保在所额定值 下都能启动 MA To check the starting torque M2 has to be compared to MA Starting is at any rate ensured because sufficient acceleration reserves are available S2 方式下允许的电机转矩为 The permissible torque of the motor for S2 operation is 4 6 NmkMM SrSr 1 51 22 电机不会过载 The motor is not overloaded 03 04 2010G tz Product Support Drive selection 4 2 设计形式为设计形式为 N 的的 Lenze 减速电机的选型减速电机的选型 Geared motor design N 求传输功率 Calculation of the process power 4 7 W nM P75 6073 60 2 22 2 求 kS2 1 4 且 Gearbox initial 0 95 时所需的电机功率 Calculation of the required motor power with kS2 1 4 and Gearbox initial 0 95 4 8 W k P P initialgearboxS req 73 4566 2 2 1 根据主电源数据选择电机电压及频率 Motor voltage and motor frequency correspond to the mains data 可选择 4 极异步电机 112 32 A 4 pole asynchronous motor is selected 112 32 电机型号 Motor size PNnrIrIA IrV frcos MrMstallMA J kW min 1 A A V Y Hz Nm Nm NM 10 3 kgm2 112 325 5144012 58 0 400 2 500 788936 5138 7105 922 8 供电电压 400V 连接方式 角接 Delta interconnection with 400 V 求减速器速比 Calculation of the setpoint gearbox ratio 4 9 4 14 2 n n i N soll 负载等级为 I Load class I is defined 由于在 S2 方式下运行 10 分钟 故每小时开关次数很少 The number of operations per hour is very small because of the S2 operation of 10 minutes 运行因子最大为 0 9 This leads to an operation factor k of max 0 9 在 G Motion const 手册中查阅 N 型减速器数据 查出 iactual Selection of iactual in the G Motion const catalogue design N 特性 Characteristics M2perm n1 IEC 连接 IEC connection iactual 14 286GST07 2N M2perm 624 Nm M2 k 若所需传递的转矩传至电机侧 则结果为 Gearbox 0 97 If the requested process torque is transformed to the motor side the result is as follows Gearbox 0 97 03 04 2011G tz Product Support Drive selection 4 10 Nm i M M Getriebeist 85 41 2 2 为校核启动转矩 必须将 M2 作为MA 为获得充足的加速裕量 必须确保在所有额定 值下都能启动 MA To check the starting torque M2 has to be compared to MA Starting is at any rate ensured because sufficient acceleration reserves are available S2 方式下允许的电机转矩为 The permissible torque of the motor for S2 operation is 4 11 NmkMM SrSr 1 51 22 电机不会过载 The motor is not overloaded 03 04 2012G tz Product Support Drive selection 4 3 制动单元制动单元 1 Braking unit 1 求转动惯量求转动惯量 Calculation of the moment of inertia 下式适于圆柱固体转动惯量的计算 For a solid cylinder the following formula applies 4 11 22 0625 0 2 kgmr m JL 从而可得 As a result the total inertia is 4 12 MotorLtotal JJJ 4 13 2 0635 0kgmJtotal 运动学分析 运动学分析 Kinematics 延迟 Delay 4 14 2 1 8 261 60 2 s dt dn dt d brake 制动时 动态传输转矩按下式计算 When braking the dynamic process torque is calculated as follows 4 15 NmJM braketotaldyn 62 16 总制动转矩 The total braking torque 4 16 NmMMM dynfriction 62 13 3 相应的制动功率峰值 The corresponding peak brake power of the process 4 17 WMP brakeozess 72 3565 3 Pr M t M1 M2 M3 M4 Meff P t PBake max PBake ave 03 04 2013G tz Product Support Drive selection 直流母线上的制动功率 The peak brake power at the DC bus is 4 18 WPPP VMotorbakeozessbrake 58 2592 Prmax 连续制动功率 Calculation of the continuous braking power 4 19 W T t PP cycle brake brakeavebrake 18 185 2 1 max 由于制动功率是连续的 因此不允许使用制动模块 必须使用制动斩波器 Due to the continuous braking power it is not possible to use a braking module The braking chopper 9352 has to be used 制动电阻最大值按下式计算 The maximum braking resistor is calculated as follows 4 20 202 58 2592 725 2 max 2 max W V P U R brake threshold brake 制动电阻最小值 Calculation of the minimum braking resistor 4 21 18 42 725 max min AI U R Chopper threshold brake 求制动电阻热容量 Calculation of the required thermal capacitance of the resistor 4 22 WstPW brakebrakebrake 29 1296 2 1 max 制动电阻值必须介于 RBrake min和 RBrake max 之间 且其热容量应大于所需制动能量 但 是 制动电阻必须满足连续制动及制动功率峰值要求 The resistor value must be between RBrake min and RBrake max and have a higher thermal capacitance than the braking energy required Moreover it must be able to handle the continuous and peak power 结论 制动电阻值 RBrake 180 As a result the following resistor can be used RBrake 180 03 04 2014G tz Product Support Drive selection 4 4 制动单元制动单元 2 Braking unit 2 驱动所需时间 Time required for one drive 制动运行时 发电模式产生的功率为 When moving downwards a generator mode power of 此时 制动功率持续上升 必须使用多台控制斩波器 The braking power arises continuously and has to be dissipated by several braking choppers 9352 制动斩波器可处理 19kW 连续制动功率 这意味着需 4 台 9352 The braking chopper 9352 can handle 19 kW continuously This means that a total of 4 braking choppers are needed Each chopper has to dissipate a quarter approx 16 kW of the total power 每台制动斩波器配备的制动电阻值为 The corresponding braking resistor per chopper is calculated as follows 又因为电阻的阻值应在 18 到 32 85 之间 同时 制动电阻应可消耗 16kW 连续制动 功率 The minimum braking resistor is 18 The selected resistor value should be between 18 and 32 85 Moreover the braking resistor must be able to handle 16 kW continuously 可行方案为将 6 支 18 电阻按下图混联 A possi ble solution is a group connection of a total of six 18 resistors 总制动电阻值为 In this case two times 3 resistors have to be connected in series and both series connections have to be set in parallel This results in the following resistor value 这样 每支制动电阻制动可消耗 3kW 连续制动功率 共计 18kW 3 kW are continuously permissible per resistor This corresponds to a total of 18 kW 结论结论 Conclusion 此例中 需 4 台 9352 制动斩波器 每台 9352 需配备 6 支各 18 的电阻组成的电阻桥 做为制动电阻 A total of 4 braking choppers 9352 are needed Each chopper receives a resistor network with six 18 resistors W s m s m kgvgmP6376505 0181 9130000sin 2 85 32 16 725 22 kW V P U P continuos threshold 18 18 18 18 18 18 min 3 18 min 3 55 m m v s t 27 183 1 183 1 1 3 1 3 1 1 RR Rtotal 03 04 2015G tz Product Support Drive selection 4 5 传动轴传动轴 Spindle 计算转动惯量计算转动惯量 Calculation of the moment of inertia 一部分转动惯量产生于传动轴的几何形状 Moment of inertia arising from the geometry of the spindle 传动轴用于圆柱体传动时 For a solid cylinder applies 4 23 SS ldJ 4 32 该固体为钢质时 For steel applies 4 24 2 4949 25 4 900 14 1012310123 kgcm mmmmlrJS 另一部分转动惯量产生于负载质量及传动轴与其轴闩之间的摩擦 Moment of inertia arising from the mass of the load and the spindle bolt with reference to the spindle 4 25 22 2 1 1500 2 cm kg h mJ gesTrans 4 26 2 38 kgcmJTrans 总转轴惯量为 As a result the total moment of inertia is as follows 4 27 STranstotal JJJ 4 28 2 25 42kgcmJtotal 03 04 2016G tz Product Support Drive selection 运动学分析 Kinematics 本例中的速度曲线如下图所示 由于该曲线在往复过程中是一致的 故在此仅对前向 运动中的数据进行计算 The diagram shows the speed profile of the application Since the profile is the same for forward and backward driving it is sufficient to examine forward driving only 求加速及延迟时间 Calculation of acceleration and delay 4 29 2 1 max 6667 0 3 0 2 0 s m s s m t v a 隐藏距离 Distance covered 4 30 ms s m tasa03 0 3 06667 0 2 1 2 1 2 2 2 1 这说明在恒定传动中存在 0 18m 的隐藏距离 传动速度为 12 m min 时 这段距离需用 0 9s This means that 0 18 m have to be covered during constant driving In case of 12 m min this takes 0 9s 4 31 s s m mm v ss t ages 9 0 2 0 03 0 224 0 2 max 2 图中各段时间为 t1 0 3 s t2 0 9 st3 0 3 st4 0 1 sT 1 6 s The individual times are t1 0 3 st2 0 9 st3 0 3 st4 0 1 sT 1 6 s 线速度与角速度的转换为 The translatory variables are transferred to the rotatory variables as follows 4 32 h v 2 4 33 h a 2 因而 角速度为 As a result the angular velocity is as follows 4 34 sm s m h v1 66 125 01 0 2 0 22 max 又因为该角速度对应转速为 n 1200 rev min This corresponds to a speed of n 1200 rev min t1t2t3t4 T v t vmax 03 04 2017G tz Product Support Drive selection 因此 该角速度为 As a result the angular velocity is as follows 4 35 2 2 19 418 01 0 6667 0 22s m sm h a 动态传递转矩如下式计算 The dynamic process torque is calculated as follows 4 36 Nm s kgmJM gesdyn 77 1 1 9 418004225 0 2 2 首先 静态转矩为 Determination of the stationary torque 4 37 N s m kggmF btotalfriction 3 29402 0 81 9 1500 2 4 38 Nm m N h FM S frictionstat 51 0 92 0 1 2 01 0 3 294 1 2 制过过程中 静态转矩可用下式求出 During the braking the following applies to the stationary torque 4 39 Nm h FM Sfrictionstat 43 0 92 0 2 01 0 3 294 2 传动轴功效可按下式计算 The spindle efficiency is calculated as follows 4 40 92 0 1137 0 01 0 1 01 0 1137 0 1 1 1 K K S S S ith 4 41 1137 0 28 10 mm mm d h K 摩擦系数取决于传动轴的类型 Lenze formula collection s 0 01 The friction coefficient of the spindle results from the spindle type Lenze formula collection s 0 01 总传递扭矩为动态转矩与静态转矩之和 The total process torque can be found by adding the dynamic and the stationary components 4 42 tMtMtM statdyntotal 其对应的功率为 The corresponding process power 4 43 ttMtP totaltotal 总传递转矩的计算值如下所示 The calculated values for the total torque are listed below 时间 Period t s 转矩 Torque Mtotal Nm t1 0 3M1 2 28 t2 0 9M2 0 51 t3 0 3M3 1 34 t4 0 1M4 0 有效转矩取决于所选电机 The effective torque is relevant for selecting a motor M t M1 M2 M3 M4 Meff t1t2t3t4 T 03 04 2018G tz Product Support Drive selection 4 44 T tMtMtMtM Meff 4 2 43 2 32 2 21 2 1 根据上表中数据 T 1 6s Meff 1 21 Nm If the values shown in the table are used with T 1 6s Meff 1 21 Nm 所选电机为 所选电机为 Selection of the motor 电源电压 230V 角型连接的 MDFMA 71 12 电机 理由 其额定转矩大于本例中所需 之有效转矩 额定转速明显大于本例中所需传递速度 MDFMA 71 12 because the rated torque is higher than the effective torque of the application The rated speed is slightly higher than the requested process speed The motor is delta connected with 230 V 电机型号 Motor type 轴高 Axis 额定转速 Speed 额定转矩 Torque 额定功率 Power 额定电流 Current 电源电压 Connection 额定频率 Frequency heightVoltage hnrMrPrIV mm min 1 Nm kW A V Hz MDXMA 71 12 7113551 80 250 85 1 5400 23050 电机型号 Motor type 功率因子 Power 功效 Efficiency 堵转转矩 Stall Torque 张紧转矩 Tightening Torque 启始电流 Starting Current 转动惯量 Moment of inertia 重量 Weight factor MrMAIA IrJM cos Nm Nm kgm2 kg MDXMA 71 12 0 700 613 43 43 80 00065 9 校核电机负载能力校核电机负载能力 Check of motor load capacity 由于电机转矩 会产生较高的动态应力 Owing to the torque of the motor the dynamic stress will be higher 4 45 Motortotalnewtotal JJJ 4 46 2 25 48kgcmJ newtotal 因此 可按下式求出新的动态转矩 The new dynamic process torque is as follows 4 47 newtotalnewdyn JM 4 48 Nm s kgmMdyn02 2 1 9 418004825 0 2 2 03 04 2019G tz Product Support Drive selection 因而 总转矩如下表所示 As a result the total torque is as follows 时间 Period t s 转矩 Torque Mtotal Nm t1 0 3M1 2 53 t2 0 9M2 0 51 t3 0 3M3 1 59 t4 0 1M4 0 有效转矩为 The effective torque is as follows Meff 1 35 Nm 在计算电流值时 应进一步校核负载能力 Load capacity can and should be examined more detailed when calculating the current 此时 应根据转矩曲线求电流曲线 In this case the current profile is calculated from the torque profile 电流 Ia r 及 If r 得自额定数据 The currents Ia r and If r result from the rated data 4 49 AAII rtotalra 05 1 7 05 1cos 4 50 AAAIII rartotalrf 07 1 05 1 5 1 222 2 根据下表的等式 可由转矩求出有效电流 Owing to the following general relation the corresponding effective current can be calculated from the torque M 4 51 ra a r I I M M 总电流为 The total current can be calculated as follows 4 52 2 2 Nfatotal III 以上小结如下