Selection and analysis of light conveyor motor case

Light conveyor belts mainly include PVC conveyor belts, rubber and plastic product conveyor belts, PU conveyor belts, etc., which are widely used in food, logistics, electronics, tobacco, wood, stone, fitness and other fields. The suitability of the conveyor motor directly affects the smooth operation and transportation of the conveyor belt. Therefore, how to choose a suitable suitable motor is an important part of the conveyor design. This article gives an example analysis of the selection and calculation of the conveyor motor.

The CD manufacturer's packaging workshop carries out engineering transformation. The project is to connect two packaging machines for production. The scheme is to use the conveyor with the robot for transportation, and the whole is controlled by PLC, supplemented by pneumatic components, positioning components, low-voltage circuit components, mechanical transmission. Components, etc., form a closed-loop control loop. The basis for selecting the conveyor motor and related calculations is as follows:

First, analyze the working conditions:

The maximum output of the SKS packaging machine is 6120 boxes per hour. The general output of the lower machine packaging machine is 70 bags per hour, that is, 7,000 boxes per 10 boxes. The conveying capacity of the conveyor must be greater than the production speed of the machine. The production speed of the conveyor is determined to be 6,500 boxes per hour, 65 bags, and the size of the CD case is 142 × 124 × 10, so the minimum conveying speed of the conveyor is 9.23 meters per hour. With a speed of 0.125 meters per minute and a conveyor drive rod diameter of Φ20mm, the conveyor drive rod speed is at least 2.45r/min.

The weight of a CD box is 70 grams, a package is 10 boxes, 700 grams, and the conveyor belt usually has 5-7 bags, so the conveyor belt has a delivery weight of 3.5kg-4.9kg and a maximum weight of 5kg. Due to the positioning of the material, the transmission distance is required to be precise, and the number of revolutions is too small. Therefore, the motor must have the following characteristics: braking and holding the load after power off; fast braking speed, the number of revolutions is too small; frequent starting can be achieved. Calculate 65 bags per hour, that is, provide 1 bag in 55 seconds, so the motor must start and stop at least 2 times per minute.

Second, the specific calculation is as follows:

1-pulley mechanism:

Total weight of belt and work ••••••• m1 = 10kg

Friction coefficient of sliding surface •••••••••••• = 0.3

Roller diameter ••••••••••• D = 20mm

Roller weight ••••••••••••• m2 = 1kg

Efficiency of belt and roller ••••••••••• = 0.9

Belt speed ••••••••• V = 28mm / s ± 10%

Motor power ••••••••••• Single phase 22050Hz

Working hours ••••••••••• 24 hours a day, 7 days a week

2 Determine the reduction ratio of the gearbox:

Output shaft speed reduction ratio: NG = (V • 60) / (π • D) = ((28 ± 14) × 60) / (π × 20) = 26.7 ± 2.7 [r / min]

Since the rated speed of the motor (4 poles) at 50 Hz is about 1500 r / min, the reduction ratio i = 60 should be selected in this range.

The reduction ratio i of the reduction gearbox is: i = (1500) / NG = (1500) / (26.7 ± 2.7) = 51 ~ 62.5

3 Calculate the required torque:

The torque required to start the belt is the maximum. First calculate the required torque at the start.

Friction force of sliding part F = ·m · g = 0.3 × 10 × 9.807 = 29.4 [N]

Load torque TL = F · D / 2 · = (29.4 × 20 × 10-3) / (2 × 0.9) = 0.326 [N · m]

This load torque is the value of the output shaft of the gearbox, so it needs to be converted to the value of the motor output shaft.

Required torque of motor output shaft TM TM = TL / i · ηG = 0.326 / (60 × 0.66) = 0.00824 [N · m] = 8.24 [mN · m] (conduction efficiency of reduction gearbox G = 0.66)

According to the use of power supply voltage fluctuations (220V ± 10%) and other considerations, the safety speed is set to 2 times. 8.24 × 2≈16.48 [mN m] Required power for motor

LPLPmnTP5.2 ~ 5.1

When the coefficient is 2, Pm = 2T 2πn = 2 × 0.01648 × 2 × π × 1500/60 = 5.17W A motor with a starting torque of 16.48 N m or more can be selected by referring to the standard motor model/performance table. Motor: 60YB06DV22, then select 60GK60H which can be matched with 60YB06DV22. 4 Check the inertia of the load inertia: Inertia of the belt and the working object Jm1 = m1 × (π × D / 2π) 2 = 5 × (π × 20 × 10-3 / 2π) 2 = 5 × 10-4 [ kg · m2 Inertia of the roller inertia Jm2 = 1/8 × m2 × D2 = 1/8 × 1 × (20 × 10-3) 2 = 0.5 × 10-4 [kg · m2] Full load inertia of the output shaft of the gearbox J = 5 × 10-4 + 0.5 × 10-4 × 2 = 6 × 10-4 [kg · m2]

Check the manufacturer's technical manual for the allowable load inertia of the dynamic output shaft 60GK60H

Jm = 0.062 × 10-4 [kg m2]. JG = Jm × i2 = 0.062 × 10-4 × 602 = 223.2 × 10-4 [kg m2]

Because J And the rated torque of the motor is selected as 40mN m, which is larger than the actual load torque, so the motor can run at a speed faster than the rated speed.

Then calculate the belt speed according to the no-load speed (about 1500r/min) to confirm whether the selected product meets the specifications.

V = (NM · π · D) / 60 · i = (1500 × π × 20) / (60 × 60) = 26.17 [mm / s] The above confirmation results can all meet the specifications.

In summary, the analysis of load conditions and load calculation are the basis for selecting the motor and gearbox. Detailed calculations can be found in the relevant sections of the Mechanical Design Guide.

Third, determine the model of the motor and related accessories.

Combined with the actual use of power supply and accessories in the factory, the electromagnetic brake motor is selected, the model is 60-YB-06D-V22 (frame size 60, electromagnetic brake motor YB, round shaft 6W, single phase 220V); The supporting gearbox model is 60-GK-60H (frame size 60, reduction gearbox 6W, reduction ratio 60, standard structure); it is directly connected to the conveyor drive rod with elastic coupling, elastic

The coupling type is 28MC08-08 (nominal outer diameter Φ28, inner diameter Φ8); the motor angle foot mounting type is RAL60.

Fourth, the conclusion.

After practical application, the motor power is selected at a moderate level, the transmission speed is suitable, the operation is stable and the failure rate is low, fully meeting the design requirements. Through the analysis of working conditions, such as design, load calculation, etc., the motor can be selected reasonably, safely and reliably. The above is only the selection and installation process. For the wiring of the control circuit, please refer to other documents.