减速机详细的选型计算及练习

1、目录(Contents)1练习简介(Brief description of the exercises)12实用工具(Aids) 23练习(Exercises)33.1结构设计形式为M的减速电机(Geared motor design M)33.2结构设计动工为N的减速电机(Geared motor design N)43.3制动单元练习1 (Braking unit 1)53.4制动单元练习2 (Braking unit 2)63.5传动轴(Spindle)74练习答案(Solutions)84.1结构设计形式为M的减速电机(Geared motor design M)84.2结构设计形式

2、为N的减速电机(Geared motor design N)104.3制动单元练习1 (Braking unit 1)124.4制动单元练习2 (Braking unit 2)144.5 传动轴(Spindle)151 练习简介(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(Brakin

3、g 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(

4、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)2 实用工具(Aids) 计算器(Pocket calculator) Lenze选型手册(Lenze catalogues) Lenze公式集(Lenze formula collection)3 练习(Exercises)3.1设计形式为M的Lenze减速电机的选型(Geared motor design M)减速电机按S2方式进行传动(运

5、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(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 H

6、z运行时间(Operating time/day):8 h所需部件(Searched components):Lenze异步电机(Lenze asynchronous motor)GST减速器(Gearbox GST)3.2 设计形式为N的Lenze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 opera

7、tion is given.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(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

8、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.)3.3 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 an

9、d braked by a servo drive as shown in the above characteristic.)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 pe

10、riod): t4 = 1 s循环周期(Cycle time): T = 7 s电机功效(Efficiency of the motor):hMotor = 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(Searched components):制动单元(Braking unit, resistor)转矩及功率曲线(Torque and power profile)3.4 制动单元2(Braking unit 2)电机(Moto

11、r): 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) j = 0应用范围:(Application:

12、 Hoist without counter-weight.)需选择(Searched components):制动单元，制动电阻(Braking unit, resistor)3.5 传动轴(Spindle)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

13、rails.)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

14、(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): mb = 0.02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, resistor04 练习答案(S

15、olutions)4.1 设计形式为M的Lenze减速电机的选型(Geared motor design M) 求传输功率(Calculation of the process power)(4.1)求kS21.4且 hGearbox, initial 0.95时所需的电机功率:(Calculation of the required motor power with kS2 = 1.4 and hGearbox, initial = 0.95)(4.2)根据主电源数据选择电机电压及频率(Motor voltage and motor frequency correspond to the m

16、ains 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 jhMrMstallMA JkWmin-1AAV Y/DHz%NmNmNM10-3 kgm2112-325.5144012.58.0- / 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 set

17、point gearbox ratio):(4.3)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 iact

18、ual 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hGearbox = 0.97(If the requested process torque is transformed to the motor side, the result is as follows: hGearbox = 0.97)(4.4)可根据电机的运行值求出C。(C could be recalculated based on the operating point o

19、f the motor.)(4.5)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)电机不会过载。(

20、The motor is not overloaded.)4.2 设计形式为N的Lenze减速电机的选型(Geared motor design N)求传输功率(Calculation of the process power)(4.7)求kS21.4且 hGearbox, initial 0.95时所需的电机功率:(Calculation of the required motor power with kS2 = 1.4 and hGearbox, initial = 0.95)(4.8)根据主电源数据选择电机电压及频率(Motor voltage and motor frequency

21、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 jhMrMstallMA JkWmin-1AAV Y/DHz%NmNmNM10-3 kgm2112-325.5144012.58.0- / 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(Cal

22、culation of the setpoint gearbox ratio):(4.9)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。(S

23、election of iactual in the G_Motion const catalogue design N.)特性 (Characteristics) ： M2perm n1 IEC连接(IEC-connection)iactual = 14.286GST07-2NM2perm = 624 Nm M2 * k若所需传递的转矩传至电机侧，则结果为hGearbox = 0.97(If the requested process torque is transformed to the motor side, the result is as follows: hGearbox = 0

24、.97)(4.10)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)电机不会过载。(The mo

25、tor is not overloaded.)4.3 制动单元1 (Braking unit 1)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)从而可得(As a result the total inertia is)(4.12)(4.13)运动学分析：(Kinematics)延迟Delay: (4.14)制动时，动态传输转矩按下式计算：(When braking, the dynamic proces

26、s torque is calculated as follows)(4.15)总制动转矩(The total braking torque) (4.16)相应的制动功率峰值：(The corresponding peak brake power of the process)(4.17)直流母线上的制动功率：(The peak brake power at the DC bus is)(4.18)连续制动功率：(Calculation of the continuous braking power)(4.19)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

27、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) 制动电阻最小值(Calculation of the minimum braking resistor)(4.21)求制动电阻热容量：(Calculation of the required thermal capaci

28、tance of the resistor)(4.22)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 powe

29、r.)结论：制动电阻值RBrake = 180 W (As a result, the following resistor can be used: RBrake = 180 W )4.4 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 di

30、ssipated 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. 16kW) of the total power. 每台制动斩波器配备的制动电阻值为：(The corresponding br

31、aking resistor per chopper is calculated as follows)又因为电阻的阻值应在18W到32.85W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 W. The selected resistor value should be between 18W and 32.85W. Moreover, the braking resistor must be able to handle 16 kW continuously.)可行方案为将6支18W电阻按下图混联。 (A possi

32、ble solution is a group connection of a total of six 18W 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 p

33、ermissible per resistor. This corresponds to a total of 18 kW.)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 W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 W resistors.)4.5 传动轴(Spindle)计算转动惯量Calculation of the moment of inertia一部

34、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)该固体为钢质时：(For steel applies)(4.24)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)(4

35、.26)总转轴惯量为：(As a result, the total moment of inertia is as follows)(4.27)(4.28)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 forwar

36、d driving only.)求加速及延迟时间：(Calculation of acceleration and delay)(4.29)隐藏距离：(Distance covered)(4.30)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)图中各段时间为： t1 = 0.3 s t2 = 0.9 st3 = 0.3

37、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)(4.33)因而，角速度为：(As a result, the angular velocity is as follows)(4.34)又因为该角速度对应转速为n = 1200 rev/min。(This

38、 corresponds to a speed of n = 1200 rev./min)因此，该角速度为：(As a result, the angular velocity is as follows:)(4.35)动态传递转矩如下式计算：(The dynamic process torque is calculated as follows)(4.36)首先，静态转矩为：(Determination of the stationary torque)(4.37)(4.38)制过过程中，静态转矩可用下式求出：(During the braking, the following applie

39、s to the stationary torque)(4.39)传动轴功效可按下式计算：(The spindle efficiency is calculated as follows)(4.40)ith (4.41)摩擦系数取决于传动轴的类型(Lenze formula collection, ms=0.01)。The friction coefficient of the spindle results from the spindle type (Lenze formula collection, ms=0.01).总传递扭矩为动态转矩与静态转矩之和。(The total proces

40、s torque can be found by adding the dynamic and the stationary components.)(4.42)其对应的功率为：(The corresponding process power)(4.43)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 Nmt1 = 0.3M1 = 2.28t2 = 0.9M2 = 0.51t3 = 0.3M3 = -1.34t4 = 0.1M4 =

41、 0有效转矩取决于所选电机。(The effective torque is relevant for selecting a motor.)(4.44)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：(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 becau

42、se 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)heightVolta

43、gehnrMrPrIV(mm)(min-1)(Nm)(kW)(A)(V)(Hz)MDXMA 71-127113551.80.250.85/1.5400/23050电机型号(Motor type)功率因子(Power)功效(Efficiency)堵转转矩（StallTorque）张紧转矩（Tightening Torque）启始电流（Starting Current）转动惯量(Moment of inertia)重量(Weight)factorMrMAIA / IrJMcos jh(Nm)(Nm)(kgm2)(kg)MDXMA 71-120.700.613.43.43.80.00065.9校核电

44、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)(4.46)因此，可按下式求出新的动态转矩：(The new dynamic process torque is as follows)(4.47)(4.48)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 Nmt1

45、 = 0.3M1 = 2.53t2 = 0.9M2 = 0.51t3 = 0.3M3 = -1.59t4 = 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)(4.50)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)总电流为：(The to