减速机详细的选型计算及练习.docx

1、减速机详细的选型计算及练习目录(Contents)1 练习简介(Brief description of the exercises) 12 实用工具(Aids) 23 练习(Exercises) 31 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(Braking unit)加

2、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(Catalogu

3、e 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(运行时间=10mi

4、n)，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 Hz运行时间(O

5、perating 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 operation i

6、s 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，作为规则连接，该型

7、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 and br

8、aked 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 period

9、): 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(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(Motor)

10、: 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:

11、 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

12、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

13、(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(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(So

14、lutions)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 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)根据主电源数据选择电机电压及频率(Motor voltage and motor frequency correspond to the ma

15、ins 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 JkWmin-1AAV Y/Hz%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 setpoin

16、t 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 iactua

17、l 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)可根据电机的运行值求出C。(C could be recalculated based on the operating point of t

18、he 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。(T

19、he 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 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)根据主电源数据选择电机电压及频率(Motor voltage and motor frequency c

20、orrespond 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 JkWmin-1AAV Y/Hz%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(Calcula

21、tion 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。(Sel

22、ection of iactual in the G_Motion const catalogue design N.)特性 (Characteristics) ： M2perm n1 IEC连接(IEC-connection)iactual = 14.286 GST07-2NM2perm = 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

23、) (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 mot

24、or 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 proc

25、ess 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。(D

26、ue 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) 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 therma

27、l capacitance 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

28、peak power.)结论：制动电阻值RBrake = 180 (As a result, the following resistor can be used: RBrake = 180 )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

29、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. 16kW) of the total power. 每台制动斩波器配备的制动电阻值为：(The correspondi

30、ng 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 possible 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