Unloaded Motors

[Gorpomon]

Hello,

I'm working on developing a sheet roller that starts out unloaded and ends up with 250lbs of sheet rolled around the shaft. I'm looking at getting a geared motor in the 25 - 40rpm range. The sheet can't move faster than the linear speed equivalent of 25-40rpm (its a 3" shaft) because its being printed on the sheet needs exposure time to dry before being rolled up.

I have 2 questions I would like some guidance on:

1. Was this calculator I used to determine full load torque correct for the application? (Weight 250lbs. Time to accelerate 10s, R1=.625ft, R2=.25ft, RPM=45) It's a hollow cylinder calculator from the website engineers edge.

http://www.engineersedge.com/motors/calculator/accelerating_torque_force_hollow_cylinder.htm

2. The machine starts out unloaded (about 10lbs for the shaft weight) and only ends at full load (250lbs), so how do I know how fast the motor will spin unloaded? Remember in my application I can't exceed the 25-40rpm speed, so it starting out really fast and then slowing down as it loads isn't an option for me. How do I figure out the starting speed and compensate for it, or size the motor for it? Or is a motor controller my only option?

Any guidance on these would be appreciated.
Thanks,

 

[desertfox]

Hi Gorpomon

You would be correct with question 1 only if you were starting with the cylinder already wound with the sheets, what you actually have is a varying inertia, but if you use a geared motor with sufficient torque capacity I can't see a problem winding the sheet on.
As the sheet winds on the cylinder the rpm will stay constant but the linear speed of the sheet will increase, the linear velocity being Angular Velocity * radius.

desertfox

 

[Gorpomon]

Desertfox, is my fundamental understanding of how the motor operates wrong? I thought that as the motor starts out unloaded, it will go faster with lower torque, and then as it loads up to the 250lbs, it will be at that torque and rpm that the motor is rated full load for.

What do you mean by a sufficient torque? Wouldn't plugging in the max conditions in that calculator be the torque the motor needs, or would it still need to be upped by some safety factor to be sufficient?

 

[desertfox]

Hi Gorpomon

What I am saying is a geared motor can produce a torque sufficient to wrap sheets on the cylinder with a constant output rpm.

desertfox

 

[waross]

Remember in my application I can't exceed the 25-40rpm speed, so it starting out really fast and then slowing down as it loads isn't an option for me.

No start out slow and then slow down.
With the most common type of motor, the three phase induction motor, there may be 2% difference between loaded speed and unloaded speed.
More suitable options may be a DC motor and drive or a Variable Frequency Drive controlled induction motor.
Inertia may only be an issue if the roll must be re-started when it is partially full. Not a problem with a properly setup drive.
This is assuming that the winder may control the sheet speed.
If the winder must cope with speeds determined by another piece of machinery, hire an expert.
Not understanding basic motor characteristics is too great a knowledge deficit to jump in to designing a sophisticated drive.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[desertfox]

Hi Gorpomon

Below is an extract from the Baldor motor site:-

Motor Characteristics
Motor Operation
As previously stated, a conductor moving through a magnetic field due to the motor action also generates a voltage which is in opposition to the applied voltage. This is the back EMF. Then for motor action the voltage equation is:


V = E + IA RA = K1 S + IA RA

where:


V = applied or terminal voltage
E = back EMF
IA = armature current
RA = armature circuit resistance's
K1 = machine constants
= flux per pole
S = speed

When comparing this equation with the voltage equation of a generator, it can be seen that in a generator the generated voltage is higher than the terminal voltage while in a motor the opposite is true. Therefore, as long as the generated voltage is less than the terminal voltage, a machine operates as a motor and takes power from the electrical side, but when the generated voltage becomes greater than the terminal voltage, the machine becomes a generator, supplies electric power, and requires mechanical energy to keep operating.

The back or counter EMF acts as a control for the amount of current needed for each mechanical load. When the mechanical load is increased, the first effect is a reduction in speed. But a reduction in speed also causes a reduction in back EMF, thus making available an increased voltage for current flow in the armature. Therefore, the current increases which in turn increases the torque. Because of this action, a very slight decrease in speed is sufficient to meet the increased torque demand. Also, the input power is regulated to the amount required for supplying the motor losses and output.

Speed Torque Curves
Speed torque curves for the three forms of excitation are shown in Figure 25. In a shunt excited motor, the change in speed is slight and, therefore, it is considered a constant speed motor. Also, the field flux is nearly constant in a shunt motor and the torque varies almost directly with armature current.

In a series motor the drop in speed with increased torque is much greater. This is due to the fact that the field flux increases with increased current, thus tending to prevent the reduction in back EMF that is being caused by the reduction in speed. The field flux varies in a series motor and the torque varies as the square of the armature current until saturation is reached. Upon reaching saturation, the curve tends to approach the straight line trend of the shunt motor. The no load speed of a series motor is usually too high for safety and, therefore, it should never be operated without sufficient load.

A compound motor has a speed torque characteristic which lies between a shunt and series motor.

Speed Regulation
Speed regulation is the change in speed with the change in load torque, other conditions being constant. A motor has good regulation if the change between the no load speed and full load speed is small.

Percent Speed Regulation = (SNL - SFL) / SFL x 100 A shunt motor has good speed regulation while a series motor has poor speed regulation. For some applications such as cranes or hoists, the series motor has an advantage since it results in the more deliberate movement of heavier loads. Also, the slowing down of the series motor is better for heavy starting loads. However, for many applications the shunt motor is preferred.

http://www.reliance.com/mtr/mtrthr.htm

If you wish to read the whole thing the link is above.

desertfox

 

[Skogsgurra]

We have been through this once before. In thread237-292765 I told you about induction motors and the use of an adjustable clutch. I first gave you an overview saying that this used to be DC territory and that there are VFDs with the needed functionality available so AC (induction motors) also can be used today.

You said that you wanted a simpler solution and I then assumed that you had understood how a constant speed motor and an adjustable clutch does work. The calculator you found is good if you have a controlled web tension and want to compensate for acceleration and deceleration. That is nothing you will ever be close to and you shall not use that calculator.

If you do not mind having a low web speed at the beginning of the reel and faster web speed when reel is full, then you do not need the adjustable clutch (could also be a mechanical speed variator).

But, if you want web speed and web tesion to be reasonable constant, you could use an adjustable clutch or a drive with the right control algoritms built-in. There are such drives for DC motors and AC motors. AC is preferred today. Especially if you are only winding (not unwinding, which usually means regeneration).

You need to make up your mind what technology to use. And to be able to do that you need to understand the differences between them. You seem to be far from that insight and I can only recommend that you get help from an experienced drives engineer.

Gunnar Englund

http://www.gke.org

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