fo winder aplication motor power

Hello

I need calculate winder motor power for metal(cooper) wire winding.

min diam:240mm
max diam:740mm
line speed 250m/min
max weight:950kg

how i calculate selct enought motor power?

pls help

best regards


Crb

crb_n said:

Hello

I need calculate winder motor power for metal(cooper) wire winding.

min diam:240mm
max diam:740mm
line speed 250m/min
max weight:950kg

how i calculate selct enought motor power? pls help best regards Crb

Click to expand...

You need to use calc HP = Torque X RPM divided by 5252.

You can get RPM from line speed. You can get an estimate of torque using a torque wrench on the equipment input shaft. Even with a calibrated torque wrench you will have some error. Know the tolerance of your tooling and add that "error" at the end of your calcs to ensure you will use an adequate sized motor.

Your motor must be able to meet the torque demanded by the load otherwise it may stall or be overloaded and have a short service life. It will not hurt the motor in the least if it is loaded to only 90% - bit inefficient I admit - but this will give a longer service life.

Dan Bentler

If this windup spool is center driven, you will need to figure maximum torque required at maximum roll diameter and maximum roll speed at minimum roll diameter. Then use the formula that Liefmotif gave you above.

The formula will produce a hp figure that will never be actually experienced but, since you need the torque in one case, and the speed in another case, you have to oversize the motor to cover the application. The same will be true of the drive that is energizing the motor.

many thaks for your advice.
my main problem, how i know torque? (how calculate)
coiling materila is cooper flate wire. Dimansion about 0.90x3.9mm. Maximum coil weight about 900kg.

thanks

Sizing winders needs a bit of thought. Winders are generally considdered a constant power application in that when you need hign speed (at the start) you only need a limited amount of torque. As the diameter increases, the torque requirement increases and the speed needed reduces.

The torque required is comprised of three separate component. The first part is tensioning torque expressed as force * distance e.g. 200N at a radius of 0.5M = 100NM. The second component is friction losses (static + dynamic) which are likely to be low in comparison to the tension torque but need to be considdered. The third component of the torque requirement is inertia; if the winder is required to accelerate and decellerate with the line then this must also be considdered.

It is also common practice to arange the gearing so as to exceed the base speed of the motor by 100% or more to achieve line speed at minimum diameter. There are two distinc advantages to this approach 1) You select a much smaller motor 2) you end up with a more favourable inertia match (reflected inertia is load inertia * the square of the gear ratio) between the motor and the load (you should aim for an inertia match of 10:1 or less).

As most of the world is metric: KW = (Torque(NM) * RPM) / 9550

Nick

Poor mans way is to fix a lever to the winding shaft somehow. Make it a metre long with a hook on it. Install a big bucket and fill it up with water. When it starts to move then you can calculate the torque from the amount of water you used. 1 litre = 1kg.

For a centre winder, it would normally be the winder that generates the tension by pulling against a speed master. Tension might be generated by a dancer roll but often the tension is generated directly by the winder motor by giving it an overspeed demand and limiting the torque to that which generates the required torque.

I'm not sure about copper but for aluminium I used to work on a value of 30N per square mm so that would be 0.9 * 3.9 * 30 = 105.3N at a radius of 0.375M = 39.48NM.

The OP will need to establish the required winding tension.

Nick

Manglemender said:

For a centre winder, it would normally be the winder that generates the tension by pulling against a speed master. Tension might be generated by a dancer roll but often the tension is generated directly by the winder motor by giving it an overspeed demand and limiting the torque to that which generates the required torque.

I'm not sure about copper but for aluminium I used to work on a value of 30N per square mm so that would be 0.9 * 3.9 * 30 = 105.3N at a radius of 0.375M = 39.48NM.

The OP will need to establish the required winding tension.

Nick

Click to expand...

Dear Nick

many thanks for your answere. ir is so helpfull.

when i calculate motor power by KW = (Torque(NM) * RPM) / 9550
rpm is for max diamater or min diam.?

many thanks for your help

crb_n said:

Dear Nick

many thanks for your answere. ir is so helpfull.

when i calculate motor power by KW = (Torque(NM) * RPM) / 9550
rpm is for max diamater or min diam.?

many thanks for your help

Click to expand...

Hi crb_n

It doesn't matter if its max or min diameter as long as the torque you've calculated and the rpm you're using are with respect to the same diameter.

You should also take note of manglemender's comments on mismatch of inertia you cannot simply calculate your power requirements and then go off and buy a motor and expect to work in a winder application. Your motor will likely need to be several sizes larger simply to keep the mismatch of inertia low, this will affect the controllability of load.

Cheers
chris

crb_n said:

when i calculate motor power by KW = (Torque(NM) * RPM) / 9550
rpm is for max diamater or min diam.?

Click to expand...

As it is a constant power application, it doesn't matter which set of circumstances you pick. Power is not the determining factor for a winder, torque is. You might calculate that you only need 5KW of power but if you then go and buy a 5KW motor, it won't work because it will not provide enough torque.

You need to calculate the torque required and then select your motor/gearbox to meet the torque requirements and the speed requirments as detailed in the previous post.

Nick

many thanks for all advice.

realy i am new for this type control.

i need any example for calculate coiler motor power for similar my aplication.
coiling material: copper
dimansion: 3.90X0.90 mm
line speed 250 m/min
mas diam:750mm
min diam:250mm
max coiling material about 900kg

gearbox :6

how much kw motor is enought for this? (ac motor)

best regards

Some motor/drive manufacturers have software that can help you size a motor for a particular task for example Siemens has a program called Sizer that is very helpful. Have you tried speaking to any motor/drive manufacturers?

Designing and controlling winders is quite a specialised job and one that I no longer do myself but I'd be happy to recommend my former employers if you wanted to get a third party involved.

Best regards,

Nick

chrisinns' comment about using both torque and speed at any given roll diameter is not correct. To size the motor, including any overspeed range as tanglemender suggests, you must use maximum required speed and maximum required torque. These do not occur at the same roll diameter.