How Should I Determine the Correct Power Rating for a Motor Winding Paper?

[tomizzo]

Hello,

I'm trying to size a motor for an application that I'm working on.

The application involves winding paper with two motors. One motor will unwind and the other motor will rewind. However, I'm trying to figure out what the motor power rating should be. Here's the worst case scenario:

Very non-elastic paper, winding to a diameter of 15 inches, at 15 lbs of tension at 50 feet per minute.

Here is my thought process, and I would like to know if anyone can pick out any incorrect assumptions.

As the paper is rewinding, the diameter will get bigger. However, because this diameter is get bigger, that means that it will take less revolutions per minutes to obtain a constant tangential speed (feet per minute of paper). In other words, as the diameter of the rewinding paper grows, the RPM of the winding motor will slow down.

Theoretically, this motor should operate at a constant power consumption. This is due to the fact that at a small winding diameter, the motor will not have to supply a lot of torque, but will have a large rpm to obtain the 50 feet per minute tangential speed. But once the diameter grows larger, the RPM will decrease but the torque has to increase since the tension force is further from the center of the motor shaft.

Using the formula that Horse Power = Ft-lbs X RPM / 5252, let's calculate the power when the roll is 15 inch diameter.

The torque required would be 15 lbs at 7.5 inches. This would equate to 9.375 foot-lbs. To achieve a tangential speed of 50 fpm at 15 inch diameter, that would be 50 /( 2 x pi x (7.5/12)). The 7.5/12 represents the radius in feet. This would equate to 12.739 RPM. Putting this into the horse power formula, this would require .02273 horse power.

But my question is, is this number fair enough to determine the power rating? If the motor is accelerating, obviously it will require torque to accelerate it. However, if the motor is moving very slowly and accelerating very slowly, this should be a small force.

So my questions are the following:

Are the assumptions I made above correct? .02273 HP seems very low.

And what would be an appropriately sized motor?

 

[CWatters]

I get roughly the same power (0.02HP) when I did the sums.

That's the power required to turn it at constant rpm. Presumably when the motor starts up it will have to accelerate up to 12rpm in a certain time? To do that it will have to overcome both the 15lbs tension and the inertia of the system (roller, roll of paper, motor, gearbox etc).

It's quite interesting to think about the other motor, the one unwinding the paper. Presumably it's load is also 15lbs of tension in the paper but assisting rotation so it's a negative load. Can you still call it a "motor" :-)

 

[tjej]

This won't answer your question directly but might help. I work for a web (ie newspaper) printer with a rewinder and only the take up roll is driven. The feed roll has a brake to regulate the tension. The tensile strength of the paper is more than enough.

When the paper is on the press it is being "pulled" through all the rollers and press units at the "folder" where all the webs come together. In our case this can be over 20 metres from the furthest roll-stand. If you look at a newspaper you can see the marks from the knurled rollers that grip the paper.

 

[tomizzo]

I get roughly the same power (0.02HP) when I did the sums.

That's the power required to turn it at constant rpm. Presumably when the motor starts up it will have to accelerate up to 12rpm in a certain time? To do that it will have to overcome both the 15lbs tension and the inertia of the system (roller, roll of paper, motor, gearbox etc).

It's quite interesting to think about the other motor, the one unwinding the paper. Presumably it's load is also 15lbs of tension in the paper but assisting rotation so it's a negative load. Can you still call it a "motor" :-)

Haha, no I would probably call that a brake. However, I'm trying to make the set up so that it will work with web heading in either direction. That means that both motors will take turns braking and leading.