Inverter Motor Help

[The Plc Kid]

Here is the data plate info.

Voltage / 230/460
HP / 30
Frame /286 T
Poles / 4
Rpm / 1765
Hz / 60
Fla / 74/37
Enc /TEBC
Duty / Cont
Type / TFP
Max Amb / 40c
Code F
Design / A
Nema Nomm Eff / 91.0
Service Factor / 1.0
Insulation Class / H1
Torque Range / 2-60 Hz
Max Safe Rpm / 3600


50% more Torque would be ideal.

Would it be best to increase hp to 50 or go with 6 pole motor

From dick dv post i understand that by going with more poles i will increase starting torque only?

I need to increase continious torque.

 

[Sparkz]

50% is not "a little more torque"! Please answer these questions first:

How is the output shaft of the gearbox connected to the conveyor? Are you able to change sprockets or pulleys in between?

Could you fit a larger motor (larger than size 286T) to the gearbox?

Would it be best to increase hp to 50 or go with 6 pole motor

You can not increase torque by changing to a 6 pole motor IF you are limited to the same motor frame size. It would be rather silly changing to a 6 pole knowing you would have to go beyond 60Hz just to compensate the loss in revs.

Here is the data plate info.

Voltage / 230/460
HP / 30
...

Surely this is one of those (infamous) 2-voltage, 9-wire motors? If so, you can almost double its power output at twice the speed if you connect the motor windings for 230V and hook it to a 60HP-460V VFD. Base frequency would have to be set to 120Hz; base voltage to 460V. In other words, the motor will be able to deliver (almost) the same torque at 3600rpm (motor power = 60HP) as it does at 1800rpm (motor power = 30HP).

 

[kamenges]

Originally posted by The PLC Kid:

From dick dv post i understand that by going with more poles i will increase starting torque only?

It depends on what you consider starting torque. For a given horsepower you will increase the motor torque in the region from zero to motor base speed. Above that the torque will decrease. So if you go from a 4 pole motor (1800 RPM synchronous speed) to a 6 pole motor (1200 RPM synchronous speed) of the same horsepower you will see an increase in torque of 50% relative to the 4-pole motor from zero to base speed of the 6-pole motor, which is probably in the area of 1175 RPM. If you then run faster the available torque will decrease until at 1.5 times base speed (about 1760 RPM) you will have the same torque as a 4-pole motor.

Keith

 

[milldrone]

Surely this is one of those (infamous) 2-voltage, 9-wire motors? If so, you can almost double its power output at twice the speed if you connect the motor windings for 230V and hook it to a 60HP-460V VFD. Base frequency would have to be set to 120Hz; base voltage to 460V. In other words, the motor will be able to deliver (almost) the same torque at 3600rpm (motor power = 60HP) as it does at 1800rpm (motor power = 30HP).

If this method is chosen, documentation is a must. Consider the poor third shift sparky that thinks he has "found the problem" when he finds the 230 V connection in the peckerhead. Also the "mid range" voltage and frequency need to be set at 230V and 60Hz

 

[DickDV]

PLC Kid, thanks for the motor nameplate data. As to your question about hp and continuous torque, kamenge's comments are exactly to the point. My question to you is: With 50% more torque, what is the maximum speed where this higher level of torque is required. Once you know that, you can use the formula hp = T x rpm/5250 to get the required motor hp.

In the examples we gave you above, using a 30hp 6 pole motor, you will have 50% more torque up to 1200rpm where you run out of hp. If you need the higher torque at higher speeds than that, you will need both a new motor and a new drive. If not, you only need the new motor.

Sparkz, I find it amusing that you refer to our marvelous dual winding dual voltage NEMA motors as "infamous"! As you point out, they can be handy for getting more hp out of a motor in a way somewhat similar to using an IEC motor to get more than nameplate kw. However, in the case at hand here, we still don't know if the power train reduction ratio can be easily changed or not. If not, your clever solution won't work. If yes, then your solution is easily the best and cheapest under the given circumstances especially if there is no room for a larger motor. Let's see what PLC Kid comes back with for a power train.

 

[The Plc Kid]

Dick

As far as power train the motors are coupled to a right angle gearbox and the output of the gearbox connected directly to the chains drive sprocket.

This chain has a constant tension system so there is no slip in that part of the drive.

I am still trying to get some gear box ratios from the manufacturer as there is nothing on the gearbox itself

 

[leitmotif]

Kid
lessee if we can summarize here.
Question is about a drying oven. You called it a stretch oven - I still have no idea of what that is.
Product moves thru oven on a conveyer chain
Chain is driven by gearbox which is driven with
30 HP motor 1800 RPM @ 60 Hz
Motor controlled with VFD for variable speed.
VFD is vector drive using encoder for speed sensing.
You have multiple identical units

You have said the problem has come up because of smaller product. Smaller product would have to travel thru oven faster to prevent "overcoooking" OR you would have to decrease the heat input.
In a couple places you have said you want more torque and in some others you said you want more speed.

I think there are a couple scenarios
ASSUMING
this conveyer is truly a constant torque type load.
For same product loading it will have same torque regardless of speed. HP is proportional to speed.
SCENARIOS
Product is lighter but needs to go thru faster to prevent burning.
Your issue at this stage for sure is speed and may also be torque depending on below.
1. Total product weight of lighter weight product is less than that of the heavier prodct.
2. Total product weight of lighter product is more because they can cram more in the oven.

In case one you need less torque (total product lighter) but more speed.
Case two demands more torque because of greater product total weight and you want more speed.

Case one you may need more HP but only because of higher speed. The lowered torque may offset the higher speed.
Case two you will definitely need more HP you are raising both torque and speed.

Suggestion find out what the torque demand of machine as loaded and operated is. Again you may be able to get a torque readout on VFD display. If that does not work out then -- you may have an exposed shaft - rig up a way to put a torque wrench on it. Ideally this will be motor shaft and you will have torque demand right then and there.
If shaft is on output of gearbox you will need reduction ratio. then divide torque reading by reduction ratio to get torque at motor shaft. Add 5 or 10% just to make sure.

To get reduction ratio turn input shaft X number of turns to turn the output shaft one turn. I am sure you know this but,, Do NOT forget lock n tag when you are around those gears and chains.

OK for options from here
1. turning down the heat is just not an option - ASSUMED
2. Speed up unit but do not put as much product in
3. Keep the product loading as is and speed unit just to prevent overcooking = more HP
4. Increase motor Hp
Crank your measured torque value thru formula
HP = Torque x RPM
--------------
5252
This will give you the HP requirement.

I cannot think of what I am probably forgetting.

I would try this out on one unit and see if works OK before doing on other units.
Dan Bentler

 

[DickDV]

OK, we have established that the power train cannot be easily changed.

That makes the "more speed" issue quite easy as long as full torque is not needed. Simply run the existing motor above 60hz. Check with the gearbox manufacturer for oil foaming issues if you need to go above 75hz.

The "more torque" issue is still on hold until PLC Kid tells us what maximum speed the extra torque needs to be at. That will set the hp requirement and tell us how much bigger the motor must be.

 

[milldrone]

As far as power train the motors are coupled to a right angle gearbox

This quote starts the alarm bells ringing in my head! Presuming this is a worm style box. Worm gear reducers are an automatic 5% to 15% reduction in output torque due to their inefficiency vs. a helical or planetary box.

 

[The Plc Kid]

My speed is ok.

I need to gain more torque and still have the same speed.

I do not think this is a worm style gear box

It is only a couple years old and is a sew eurodrive brand.

Right now we run full speed 1800 rpm and that does not deliver enough torque on our heaviest product.

I would like to stay close to the same physical speed ( gear box Output) and gain 30-50% more torque.

Would increasing hp to 50 from 30 at the same rpm give me the additional torque at the same speed?

The heavy product runs slower that is why the oem suggested getting a lower rpm motor to provide more torque at the heavy products low speed but if i understand correctly when we overspeed to run the light products at high line speed the the torque will fade and may not be adequate for the light products.

Am i on track?

I have a little room to go with a slightly physically larger motor.

 

[leitmotif]

My speed is ok.

I need to gain more torque and still have the same speed.

Right now we run full speed 1800 rpm and that does not deliver enough torque on our heaviest product.

I would like to stay close to the same physical speed ( gear box Output) and gain 30-50% more torque.

Would increasing hp to 50 from 30 at the same rpm give me the additional torque at the same speed?

The heavy product runs slower that is why the oem suggested getting a lower rpm motor to provide more torque at the heavy products low speed but if i understand correctly when we overspeed to run the light products at high line speed the the torque will fade and may not be adequate for the light products.

Am i on track?

I have a little room to go with a slightly physically larger motor.

Kid
What are motor speed (or VFD output freq) with the heavier and teh lighter products.

You have been given the appropriate formula
HP = T x RPM/5252 by myself and several others. you have not said you do not understand this so I must assume you do.
now crank in 50 and 30 HP solve for T and give us the answers.

Dan Bentler

 

[milldrone]

I need to gain more torque and still have the same speed.

This implies you need more Hp.

Right now we run full speed 1800 rpm and that does not deliver enough torque on our heaviest product.

Let me ask the question in a different way. You say that your heaviest product needs more torque. At what speed does this heavy torque requirement occur? If this happens only at low speeds at what speed do you have more than enough torque?

Now days with VFD's being so popular a common way to design something that has diminishing torque requirements is to set up (mechanically) the top speed at 90 Hz. I think this is what the OEM is saying.

 

[mordred]

I could be off the cuff here but I was looking at the specifications of the Powerflex drive and he may have some other options in that there are several parameters that control torque % with feedback or without feedback
look like this drive may allow tighter torque control with configuration changes
http://literature.rockwellautomation.com/idc/groups/literature/documents/um/20d-um006_-en-p.pdf

I've posted a link to the manual I'm referring to
might be an idea if you talk to your AB distributer and get their recommendations as well as those here

 

[Sparkz]

Would you be able to fit a motor frame size 324T to the gearbox? My guess is no but if I'm wrong, you would have 2 options:

1. Fit a 40HP-1800rpm-230/460V motor. Replace the drive by a 40HP-460V type. Torque will rise 33.3% and will be available from 0 to 1800rpm. Any type of product would benefit and can be transported faster if required. Most likely, the heavier product could also be transported at 1800rpm.

2. Fit a 30HP-1200rpm-230/460V motor. No need to replace the drive. Torque will rise 50% and but will only be available from 0 to 1200rpm. While the heavier product is not transported at top speed, maybe 1200rpm is enough to do the job. You can still reach the original top speed by overrevving the motor to 90Hz and still have (almost) the same amount of torque you had before the makeover. So the lighter product can also be transported at its original speed.

I just keep wondering why the original 30HP motor is no longer adequate...

 

[mordred]

360 Min Flux​

Sets the smallest level of flux used to convert​

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Par 303 [Motor Torque Ref] to a current​

reference above base speed. Note: Changed the minimum value from “0.2500” to

this is one of those parameters I was referring. Do you have a copy of drive explorer if so can you post a copy of the parameters of your drive?