Torque and motor sizing for an idiot

Hi all, I have spent about 3 days after hours searching around the net and I have picked up enough information to order a 3kW 8 pole for an AC vector drive winder project, but I am hoping someone here may be able to provide the holy grail of torque and motor sizing for bubba.
Anything would help as it all adds to the learning curve and maybe someone in the future who adds "Torque", "Motor Sizing" and "Idiot" as a search string.
Thanks
Nathan

Please explain how you came up with a 3kW 8-pole motor to start with.

Do you mean 8 pole = 4-pole pairs?

You need to start out with providing more information.
First, ignoring the motor itself, you need to know the requirements of the winder. Start with these (adapt to your units):

PLI - Pounds per linear inch tension.
Maximum Web Width.
Core OD - Core diameter.
Maximum OD - Maximum finish diameter.
FPM Max - Maximum Speed of product.
Acceleration/Deceleration requirements.
Taper Requirements.

The basics are that you want a motor/gearbox combination that can develop both the maximum speed at core, as well as have enough shaft torque to deliver the required tension to the web at the maximum diameter.

Maximum tension required is PLI * Maximum Width.
Minimum required shaft torque is the maximum tension divided by the maximum diameter.

As a general rule, I multiply my maximum shaft torque by 1.35 to account for the acceleration requirements and losses.

Well this is what I cam up with...

I (Mass monment of inertia (worse case)) = 0.5 x 400kg x 0.25^2 = 12.5 kg m^2

Acceleration (x) worst case on surface winder drum is 0 - 106 rpm in 20 seconds = 33 rad/s

T = IX = 12.5 x 33 = 416 Nm

Web tension max (overkilled) = 1200N
Surface winder drum diameter = 240mm
T = f x d = 1200 x 0.120 = 144Nm

416Nm + 144Nm = 560Nm

3kW 8 pole @ 715 rpm devolops 52.7Nm. Use 10:1 gearbox = 527Nm

At max line speed rpm of motor = 1060rpm

So, how bad is it? Please be gentle! I'm merely the sole plant Sparky here looking to broading m knowledge.

Basically, for any application including center-driven winders, you must select a power train and motor that will provide the speed range and torque that the load requires. A quick look-thru of your calculations tells me that you have done this.

I note with interest your choice to run the motor overspeed from 750rpm to 1060rpm. That's a very good way to optimize a winder design. On small motors like this, you can easily go to 150% speed as long as there is enough torque left to run the load.

I don't have the depth of experience on winders as some others on this BBS so they may have other better insights.

But, my first reaction is congratulations! You've done a good job on a rather tricky application.

This is a suface winder, yes ?

I've always found this to be a very useful reference:

http://www.eaton.com/ecm/groups/public/@pub/@eaton/@ee/documents/content/tb04000002e.pdf

Ah!!!!!!! Good catch there, LD. I went on believing this was a center-driven winder. If it's a surface-driven winder, then the calculations and my conclusions are all wrong.

HJTRBO, please clarify what type of winder this is.

Yes it's a surface winder.

I think I can see a mistake too. I haven't included the final roll diameter. As the final roll and the surface wind drum are actually a ratio. (Finished roll diamter / Surface winder drum)

So 500 / 240 = 2.08:1

So my reflected torque at the winder drum is 416 / 2.08 = 200Nm

200Nm + 144Nm = 344Nm

Am I on the right track?

Tom, excellent link, thank you.

What Inertia are you calculating (=12.5) and where is it referred to ?

A diagram is essential.

OK, now that I'm on the surface winder page, let's see. First, I find the continuous torque rating of an 8 pole 3kw motor to be 40Nm ( 3 x 9549/715 = 40). But, your choice of motor probably is still ok since you can use motor short-term torque for the acceleration torque.

At max motor speed of 1060rpm, the continuous torque has fallen to 40 x 715/1060 = 27Nm but that is still more than the running torque. The short-term torque, assuming the motor is designed for 200% s.t. torque would reduce to 40 x 2 x 715/1060 x 715/1060 = 36.4Nm. (Short-term overload torque reduces by the overspeed ratio squared)

That 364Nm on the output of the gearbox is short of the 560Nm by quite a bit but, since most of the 560Nm is accel torque, I think you will be ok.

Be sure to size the drive short-term amps to be at least 200% of motor FLA or you won't get all the overload torque the motor is capable of. Also, the motor will need an auxiliary blower for low speed cooling and you will need to consider how fast you need to stop. If you are using the drive/motor system for stopping purposes, you will need to recheck both the motor and the drive for stopping torque capacity. It is not uncommon for the braking conditions to require a larger motor than the motoring conditions.

Hope this gets you going in good shape. And, how about being done with that "idiot" bit. You sound like a pretty solid person and I applaud you for trying to think this through and then being willing to suffer through a review of your work to be sure its ok.

DickDV thanks for the kind words. I haven't even thought about stopping the darn thing yet. Especially if and when the operator hits the fast line stop! Small steps...

LD I am using Inertia for a cylinder equation. I = 0.5mr^2. The roll once finished typically will have a mass of 400kg and a diameter of 500mm.

I'll get a drawing done when I get to work

Is the inertia of the winding drum negligible ?

Here is a concept overview of what I am desigining.