Selection And Case Analysis Of Light Conveyor Motor

- Jul 12, 2019-

Light conveyor belts mainly include PVC conveyor belts, rubber and plastic products 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 belt motor directly affects the smooth running and conveying capacity of the conveyor belt. Therefore, how to choose a suitable matching motor is an important part of the conveyor belt design. This paper gives an example analysis of the selection and calculation of the conveyor belt motor. .

       A CD manufacturer's packaging workshop carries out technical transformation. The project is to connect two packaging machines for production. The scheme is to use conveyor belts with robots to transport, 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 the selection of the conveyor motor and related calculations are as follows:

First, the analysis of 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, which is 7,000 boxes per 10 boxes. The conveying capacity of the conveyor belt should be greater than the production speed of the machine. The production speed of the conveyor belt 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 belt is 9.23 meters per hour. At 0.154 meters per minute, and the conveyor belt drive rod diameter is Φ20mm, the conveyor belt drive stick speed is at least 2.45r/min.

The weight of a box of CDs is 70 grams, one package is 10 boxes, 700 grams, and the conveyor belts generally have 5-7 bags, so the conveyor belt has a delivery weight of 3.5KG-4.9KG, and the maximum weight is 5KG. Due to the material positioning, the conveying distance is required to be accurate, and the over-rotation amount is small. Therefore, the motor must have the following characteristics: braking and holding the load after the power is cut off; braking speed is fast, the amount of over-rotation is small; frequent starting can be achieved. Calculate 65 bags per hour, that is, deliver 1 bag for 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:


 

The total weight of the belt and the work••••••• m1=10kg

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

The diameter of the roller ••••••••••••D=20mm

The weight of the roller ••••••••••••• m2=1kg

Belt and roller efficiency •••••••••• η = 0.9

Belt speed •••••••• V = 28mm/s±10%

Motor Power••••••••••• Single Phase 220V50Hz

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

2 Determine the reduction ratio of the gearbox:

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

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

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

3 Calculate the necessary torque:

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

   Friction force of the sliding portion F=μm·g=0.3×10×9.807=29.4[N]

Load torque TL=F·D/2·η=(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.

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

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

LPLPmnTP5.2~5.1

When the coefficient is 2, Pm=2T·2πn=2×0.01648×2×π×1500/60=5.17W The 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 choose 60GK60H which can be combined with 60YB06DV22. 4Check 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 The inertia of the roller inertia Jm2=1/8×m2×D2 =1/8×1×(20×10-3)2 =0.5×10-4[kg·m2] The 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 60GK60H motor output shaft

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

Since J<JG, that is, the load inertia inertia is less than the allowable value, it can be used. And the rated torque of the selected motor is 40mN·m, which is larger than the actual load torque, so the motor can run at a faster speed than the rated speed.

    Then calculate the speed of the belt according to the speed at no load (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 are all able to meet the specifications.

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

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

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

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

Fourth, the conclusion.

After the actual application, the motor power is selected moderately, the conveying speed is suitable, the operation is stable, and the failure rate is low, which fully meets the design requirements. Through the analysis of working conditions, design examples, load calculations, etc., the motor can be selected reasonably, safely and reliably. The above is only the process of selection and installation. For the wiring of the control circuit, please refer to other literature.

Light conveyor belt motor