LEROY SOMER engine sizing

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Hi all

I'm back for a new question but it still concerns the same problem: as part of an end-of-studies project, I'm asked to size a motor (Leroy Somer three-phase motor imposed) for a trolley tipper. I am attaching an image of the project so that you can get an idea.

I start with the data and my calculations:

The total mass (trolley + raw material + stainless steel tubes) is 3100 N

The carriage will tilt in 10 seconds by 135° or 3/4 of a Pi, the distance to our furthest point is r=0.95m; the speed is 2.25rpm or 0.224m/s; the angular velocity is  w=0.2358 rad/s; The torque C= (FxV)/w   i.e. C= (2945x0.224)/0.2358= 2798Nm; Acceleration a=0.0528m/s.

Finally I calculate the power I need: - I don't know if it is necessary to calculate the acceleration power as follows: Pa= M x a x Vmax = 3100 x 0.0528 x 0.224 = 36.67 W

- Power Requirements: P=C*w=2798*0.2358=670W ; I apply an efficiency of 0.5 (?) i.e. 670/0.5 = 1340 W

So Total Power = 1340 + 36.67 = (approximately) 1377 W

So I know that you need a motor with a minimum power of 1377 W (1500 W I guess). From there, I don't know how to choose the engine I need because it's quite vague. I spotted geared motors that seem to be suitable at LS, I'm attaching a screenshot:

The image is not very sharp, I hope you will be able to read, I also attach the link https://www.leroy-somer.com/documentation_pdf/3733_fr.pdf

What I like to understand is how, with the results I found, I accurately determine the engine I need?

In the image above, they indicate an output speed and a reduction. Is this speed the speed I will get? So in the end, the Reduction Index doesn't really concern me? They also talk about Nominal Release Moment... What is this info for, how to use it??

There you go, I know that's a lot of questions, but if some of you would help me see more clearly, you should know that it would have been very much appreciated. And please excuse me for my lack of knowledge.

Best regards, and have a good day to all.

Fabrice


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Hello

As you are in the case of an end-of-study project, put all the information you have and that you have calculated or can calculate.

I graciously offer you a few points of knowledge   ;-)  ;-) which is one of the privileges of veterans.    ;-)

For the acceleration speed you have to consider several points , the most important of which is the position of the CG as I suggested earlier.

Remember that the center of rotation is almost in the upper right corner (and not in the middle of the carriage. This means that the effort will increase very gradually between 0 and 35° so the load asked of the engine is not the maximum effort from the start.

Be careful not to make a mistake with this acceleration story for two reasons.

  1. As you have a geared motor, it is the same as spreading the torque over time (remember that input torque = output torque but spread over time). So before having made a quarter turn on the output shaft the engine will have made several dozen or more).
  2.  A very important point concerns the non-reversibility of motion.
    QUéZACO?
    If you take a screw reducer, even if you  stop the power supply during the climb, well your load is not likely to go down. This is a question of safety because if the operator presses the emergency stop button, the truck will remain completely stationary under the load. So take a model where you have a non-reversibility because 310kg that goes down it breaks any limb and even worse if you have a nutcracker effect somewhere.
  3. If you take a worm screw model, there is no noticeable acceleration.

When you are in the down position, the CG  is closer to the axis than when you are at 90°, so the maximum torque will be in this zone and not at the start, so it would favor acceleration if you were using a geared motor with a low reduction.

So to check your max torque, it's equal to the CG lever arm between 45° and 80°. this gives you the maximum torque to bring back in Nm at the output shaft (plus a safety margin of 25 to 50% depending on the geared motor chosen.
What you should be concerned about is the torque because the power in Watt you will be confirmed by the supplier's chart.
Focus on pages 14 to 15 and especially on the maximum allowable time MMax

The slight oversizing is to be taken into account for jerks when stopping the trolley when the product is being dumped or the fact that the operator will deliberately jerks by going back and forth with the motor to bring the material down.

So you can't say that the reduction index doesn't concern you because the greater the reduction, the less effort you will have at start-up, so, instantaneous acceleration and the greater the irreversibility will be and the greater the resistance to jolts (which are formidable) because they can multiply the force by holes or four on the shaft. So rather a screw reducer that absorbs shocks better by construction.

Kind regards

PS:  the ATEX Variable is not to be taken into account in your choice, but the waterproofing is quite good yes (minimum IP 45) because of the water jets in this industry.

 

 

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Hello Zozo,

Once again, a big thank you for all this information!

I take good note of it, I will be able to move forward more serenely.

Have a nice day

Kind regards

Fabrice

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