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) (1)

2 实用工具(Aids) (2)

3 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(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 练习简介(Brief description of the exercises)

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=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): V Mains = 400 V

主电源频率(Mains frequency): f Mains = 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)

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 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): V Mains = 400 V

主电源频率(Mains frequency): f Mains = 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 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): M Friction =3 Nm 最大速度(Max. speed):

n = 2500 rpm 加速时间(Acceleration time): t 1 = 2 s 延迟时间(Delay time): t 3 = 1 s 静止周期(Rest period): t 4 = 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): J Motor = 10 kgcm 2 变频器功耗(Power loss of the inverter): P V = 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)

t1t2

t3t4

T

n

t

3.4 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(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 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(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, resistor0

4 练习答案(Solutions)

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)

W n M P 75.607360

22

22=???=

π

(4.1)

求k S2＝1.4且 ηGearbox, initial ＝0.95时所需的电机功率:

(Calculation of the required motor power with k S2 = 1.4 and ηGearbox, initial = 0.95)

W

k P P initial

gearbox S req 73.4566,22

,1=?=

η (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 mains

data.)

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):

4.142

==

n n i N

soll (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i actual (Selection of i actual 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ηGearbox = 0.97

(If the requested process torque is transformed to the motor side, the result is as follows: ηGearbox = 0.97)

Nm i M M Getriebe

ist 85.412*2=?=

η (4.4)

可根据电机的运行值求出C 。(C could be recalculated based on the operating point of the motor.) 13.1*

2

=?

=M M c c N

new (4.5)

为校核启动转矩，必须将M 2* 作为M A ，为获得充足的加速裕量，必须确保在所额定值下都能启动：(M A .To check the starting torque, M 2* has to be compared to M A . 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 S 2-operation is)

Nm k M M S r S r 1.5122,=?=

(4.6)

电机不会过载。(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)

W n M P 75.607360

22

22=???=

π

(4.7)

求k S2＝1.4且 ηGearbox, initial ＝0.95时所需的电机功率:

(Calculation of the required motor power with k S2 = 1.4 and ηGearbox, initial = 0.95)

W

k P P initial

gearbox S req 73.4566,22

,1=?=

η (4.8)

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.)

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):

4.142

==

n n i N

soll (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i actual 。(Selection of i actual in the

G_Motion const catalogue design N.)

特性 (Characteristics) ：

? M 2perm ? n 1

?

IEC 连接(IEC-connection)

i actual = 14.286

GST07-2N

M 2perm = 624 Nm ≥ M 2 * k

若所需传递的转矩传至电机侧，则结果为ηGearbox = 0.97

Nm i M M Getriebe

ist 85.412*2=?=

η (4.10)

为校核启动转矩，必须将M 2* 作为M A ，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：(M A .To check the starting torque, M 2* has to be compared to M A . 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 S 2-operation is)

Nm k M M S r S r 1.5122,=?=

(4.11)

电机不会过载。(The motor 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) ：

22

0625.02

kgm r m J L =?=

(4.11)

从而可得(As a result the total inertia is) Motor L total J J J +=

(4.12)

20635.0kgm J total =

(4.13)

运动学分析：(Kinematics) 延迟Delay:

21

8.261602s

dt dn dt d brake =???==

πωα (4.14)

制动时，动态传输转矩按下式计算：(When braking, the dynamic process torque is calculated as

follows)

Nm J M brake total dyn 62.16=?=α

(4.15)

总制动转矩(The total braking torque)

Nm M M M dyn friction 62.133=-=

(4.16)

相应的制动功率峰值：(The corresponding peak brake power of the process) M

t

M 1

M 2

M 3

M 4

M eff

P

t

P Bake,max

P Bake,ave

直流母线上的制动功率：(The peak brake power at the DC bus is)

W P P P V Motor bake ozess brake 58.2592

,Pr max ,=-?=η (4.18)

连续制动功率：(Calculation of the continuous braking power)

W T t P P cycle

brake brake ave brake 18.18521

max ,,=??=

(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 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)

()Ω==≤20258.25927252

max ,2

max ,W

V P U R brake threshold brake

(4.20)

制动电阻最小值(Calculation of the minimum braking resistor)

Ω≈=≥

1842725

max,min ,A

I U R Chopper threshold brake

(4.21)

求制动电阻热容量：(Calculation of the required thermal capacitance of the resistor)

W s t P W brake brake brake 29.12962

1

max ,=??=

(4.22)

制动电阻值必须介于R Brake,min 和R Brake,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

R Brake,min and R Brake,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R Brake = 180 Ω (As a result, the following resistor can be used: R Brake = 180 Ω )

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 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 Ω到32.85 Ω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 Ω. The selected resistor value should be between 18 Ω and

32.85 Ω. Moreover, the braking resistor must be able to handle 16 kW continuously.)

可行方案为将6支18 Ω电阻按下图混联。 (A possi‰bl e solution is a group connection of a total of six 18 Ω 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(Conclusion)

此例中，需4台9352制动斩波器，每台9352需配备6支各18 Ω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 Ω resistors.)

W

s m s m kg v g m P 6376505,0181

.9130000sin 2

=???=???=?()Ω

===85.321672522

kW

V P U P continuos threshold 18 Ω18 Ω18 Ω18 Ω

18 Ω18 Ωmin

3.18min

355===m m v s t Ω

=Ω

?+

Ω?=

?+?=

271831

18311

31311R

R R total

4.5 传动轴(Spindle)

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)

ρπ

???=

S S l d J 432

(4.23)

该固体为钢质时：(For steel applies)

2

494925.4900)14(1012310123kgcm

mm mm l r J S =???=???=-- (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)

2

22115002??

?

????=??? ????=ππcm kg h m J ges Trans

(4.25) 238kgcm J Trans =

(4.26)

总转轴惯量为：(As a result, the total moment of inertia is as follows) S Trans total J J J +=

(4.27)

225.42kgcm J total =

(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 forward driving only.)

求加速及延迟时间：(Calculation of acceleration and delay)

2

1

max 6667.03.02

.0s m s s m

t v a ==

= (4.29)

隐藏距离：(Distance covered)

()m s s

m t a s a 03.03.06667.021212

221=??=??=

(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.)

s s

m m

m v s s t a

ges 9.02

.003.0224.02max

2=?-=

?-=

(4.31)

图中各段时间为： t 1 = 0.3 s t 2 = 0.9 s

t 3 = 0.3 s t 4 = 0.1 s T = 1.6 s 。

(The individual times are: t 1 = 0.3 s t 2 = 0.9 s

t 3 = 0.3 s t 4 = 0.1 s T = 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) h

v

??=πω2

(4.32)

h

a ??=πα2

(4.33)

因而，角速度为：(As a result, the angular velocity is as follows)

s

m s m

h v 166.12501.02

.022max =??=??=ππω (4.34)

t1t2

t3t4

T

v

t

v max

因此，该角速度为：(As a result, the angular velocity is as follows:)

22

/19,41801,0/6667,022s m

s m h a =??=??=ππα

(4.35)

动态传递转矩如下式计算：(The dynamic process torque is calculated as follows)

Nm s kgm J M ges dyn 77.11

9.418004225.022=?=?=α (4.36)

首先，静态转矩为：(Determination of the stationary torque ) N s m

kg g m F b total friction 3.29402.081

.91500=??=??=μ (4.37)

Nm m N h F M S friction stat 51.092

.01201.03.29412=???=???

=πηπ (4.38)

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)

Nm h F M S friction stat 43.092.0201

,03.2942=???=???

=π

ηπ (4.39)

传动轴功效可按下式计算：(The spindle efficiency is calculated as follows)

92.01137

.001.0101

.01137.0111=+

?-=

+

?-=

K

K S

S S μμη

(4.40)

ith 1137.02810=?=?=

mm

mm

d h K ππ (4.41)

摩擦系数取决于传动轴的类型(Lenze formula collection, μs =0.01)。[The friction coefficient of

the spindle results from the spindle type (Lenze formula collection, μs =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.)

)()()(t M t M t M stat dyn total +=

(4.42)

其对应的功率为：(The corresponding process power)

)()()(t t M t P total total ω?=

(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)

M

t

M 1

M 2

M 3

M 4

M eff

t1

t2

t3

t4

T

T

t M t M t M t M M eff

4

2

4323222121?+?+?+?=

(4.44)

根据上表中数据：T = 1.6s ； M eff = 1.21 Nm 。(If the values shown in the table are used with T =

1.6s: M eff = 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 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.)

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.)

Motor total new total J J J +=,

(4.45)

2,25.48kgcm J new total =

(4.46)

因此，可按下式求出新的动态转矩：(The new dynamic process torque is as follows) α?=new total new dyn J M ,,

(4.47)

Nm s kgm M dyn 02.21

9

.418004825.02

2=?= (4.48)

因而，总转矩如下表所示：(As a result, the total torque is as follows)

有效转矩为：(The effective torque is as follows)

M eff = 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I a,r 及I f,r 得自额定数据。(The currents I a,r and I f,r result from the rated data) A A I I r total r a 05.17.05.1cos ,,=?=?=?

(4.49)

()()A A A I I I r a r total r f 07.105.15.12

22

,2,,=-=

-=

(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.)

r

a a r I I M M

,= (4.51)

总电流为：(The total current can be calculated as follows)

2,2

N

f a total I I I += (4.52)

电机有效电流I total,eff = 1.33 A 。(The effective motor current is calculated with I total,eff = 1.33 A.) 电机未过载。(The motor is not overloaded.)

选择变频器Selection of the frequency inverter:

首先，可根据电机额定功率进行粗选。(When selecting the frequency inverter, it is possible to make

a rough selection based on the rated motor power.)

但是，应仔细计算电机平均电流(算术平均值)。[The mean motor current (arithmetic mean value)

has, however, to be calculated in detail.]

A T

t I t I t I t I T

t I I total total total total n

i i

i total ave 3.14

4,33,22,11,1

,=?+?+?+?=

?=

∑

= (4.53)

斩波频率为8 kHz 时，选E82EV 251_2B 就足够了，且不会造成过载。( With a chopper

frequency of 8 kHz, the vector E82EV 251_2B is sufficient. The max. current in not exceeded either.)

求制动电阻：(Calculation of the braking resistor)

制动晶体管内置于变频器中，因此，仅有一个电阻的阻值必须加以计算。(A braking

transistor is already integrated in the frequency inverter. Therefore, only a suitable resistor has to be calculated.)

制动功率峰值为：(The peak braking power is)

W

W s Nm P M P FU

V Motor brake brake 963063.0/166.12559.1,max max ,=-???=-??=ηω (4.54)

制动电阻的最大阻值为：(The maximum braking resistor can be calculated as follows)

()Ω==≤1464963752max ,2

max

,W

V P U R brake threshold brake (4.55)

制动电阻最小阻值为：(Calculation of the minimum braking resistor)

Ω==≥

44185.0375

max,min ,A

I U R Chopper threshold brake

(4.56)

制动电阻热容量为：(Calculation of the thermal resistor capacity required)

W s s W t P W brake brake brake 4.143.0962

1

21max ,=??=??=

(4.57)

连续制动功率为：(Calculation of the continuous braking power)

W s

s

W T t P P cycle brake brake ave brake 96.13.0962121max ,,=??=??=

(4.58)

以上计算说明，可选择470 Ω电阻作为制动电阻。(The calculated values show that the 470 Ω

resistor seems to be suitable for this application.)