High cycle lift motor 2

[Barry1961]

A customer has a lift that is loading/unloading racks of parts. The rack has 50 shelves which are unloaded in 12 minutes. An example single unload cycle will consist of, starting from out feed, high speed up, low speed up, off/brake 1 sec, low speed up 1 sec, off brake 2 sec(unload shelf), low speed down 1 sec, off/brake 1 sec, high speed down, low speed down 1 sec, off brake 2 sec(out feed parts). The high speed time varies on distance to shelf, 14.4 seconds average cycle time.

The drive consists of two 2/12 pole motors feeding into 4.2:1 inline reducers which then each feed into a double input 15:1 worm reducer. One 2/12 motor is for lift up and the other for lift down. The motors are control by motor starters/relays, no electrical soft start. The free wheeling motor which is being driven through the inline reducer/worm/inline increaser(back driven reducer) acts as a flywheel which gives a mechanical slow start. The high speed starting and stopping of the flywheel seems to be a significant part of the energy being used.

Motor is 3ph, 480V, 2/12
3.5kw/4.7hp, 3458 rpm, 6.6 amp
.52kw/.7hp, 515 rpm, 3.2 amp, 2.6 amp measured

The drive sprocket for lifting has a 4.8” pitch diameter, 2.4” radius.

Problems:
1. Motors burning out, both.
2. Gearboxes failing, all of them.
3. Brake will last up to a year with proper maintenance.
4. No budget or time to install a servo drive.

I am looking at replacing the current drive with a single 5 hp and Vector Drive. The horsepower I come up with to accelerate the load a .25g/100 in-sec^2 is between 1.5 and 2 horsepower. With the single motor I will be doing twice the high speed starts, 8.3 starts per minute but it will be soft starts.

Questions:
1. Can a 5 hp motor start and stop a 2 hp load 8.3 times per minute?
2. The measured low speed current of 2.6 amps with a FLA of 3.2 amps translates into what percent of full load torque? A FLA of 3.2 seems high to me for a .7 hp motor so the is something going on I don’t understand, probably a lot. I am trying to calculate the carriage weight from the measured amp draw.

Barry1961

 

[Skogsgurra]

No wonder the motors burn out. And no wonder the gears don't last. That is the worst design I have ever heard of. I guess that most of the energy is used to back drive that idle "fly-wheel" motor. Why has it been designed like that? Why not one motor? Was it too difficult to reverse a two-speed motor? And the idea of making a soft start by using a fly-wheel is just hilarious. Rube Goldberg?

Re your questions: Yes, I have no doubt at all. Starting via a VFD is very nice to the motor as long as the ramp is at least one or two seconds. There might be some concern about the brake. Many inverters have built-in brake control that work very well.
The low speed is equal to having a 12 pole motor. They usually have a rather bad power factor. I do not think that you can calculate weight from motor current. If your 2-pole motor can lift the carriage - so should your VFD fed motor also. If the vector drive is a good one, you should be able to have full torque almost down to zero speed. For real hoisting performance, you may need to add an encoder.


Gunnar Englund

http://www.gke.org

 

[Barry1961]

This is probably a very stupid question but I just want to make sure. This is a strange motor to me.

With, .52kw/.7hp, 515 rpm, 3.2 amp, at FLA of 3.2 amp this would be 86 in-lb of torque?

Care to guess at percent of full load torque at 2.6 amp? I am going to size everything with a service factor of 3 due to the high cycle but am still curious.

This is a Italian machine with Bonfiglioli. They have a bunch of these in a plant in USA. Maybe the designer is getting a precentage of the after market sales.

Barry1961

 

[Skogsgurra]

Italians. Nice food and wine! Sometimes very special solutions...

Sorry, I don't do inch-pounds - too complicated. But I do SI. 520 W at 515 RPM (or 54 rad/sec) is 9.6 Nm.



Gunnar Englund

http://www.gke.org

 

[Skogsgurra]

BTW: PF and efficiency taken together seem to be around 20 percent (yes 20 %). That means that your current measurement won't say much about the actual load.

Gunnar Englund

http://www.gke.org

 

[waross]

Hello Barry1961 and skogsgurra;
Try this link to the Cowern Papers on the Baldor Motor site. The paper on RMS motor loading may be useful to you.

http://www.baldor.com/pdf/literature/PR2525.pdf

respectfully

 

[Barry1961]

Thank you both for the information even though it was not the answer I wanted. My brain has now exceeded the maximum safe storage capacity so I will have to do a wet erase this weekend to free up some memory.

The more I learn the less I know.

Thanks again!!

Barry1961