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Is it possible to calculate motor running speed?

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MechSteve

Mechanical
Sep 7, 2005
4
US
Hi, I feel silly but I am completely stumped by this. Here is the situation:

I have a motor shaft attached to a disk coated in rubber, friction-turning an adjacent larger cylinder. Both are pointed vertically. A load is balanced directly on top of the cylinder and is being spun (not lifted) by the setup.

The motor we purchased was not powerful enough to spin the load. I am attempting to calculate the REQUIRED TORQUE to spin the load at a CERTAIN RPM. But I can't find the proper equation. Help????????????




(I can estimate friction and weight, desired rpm, all specs on the motor. The problem as I see it is all the equations I've used call for constant acceleration, but I am looking for the speed at which the motor is at equilibrium with opposing forces and not speeding up anymore- the acceleration has asymptoted to zero. I don't even know what factor causes a motor to reach its final speed, unless it is soley internal motor specs)
 
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I’m not sure I understand your setup, but...
If you are using a NEMA Design B squirrel cage induction motor within it’s torque limits and with a fixed frequency supply at nameplate voltage, the motor should will accelerate to a speed within 5% of it’s nameplate speed.


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Hi MechSteve

I think I understand Your difficulty and I have tried to put
together an energy balance to try and help you out.

workdone by a constant Torque = T*[θ]

and T*[θ] = 0.5*I*[ω]^[²]

so Torque required to turn cylinder
at a given speed:-

T= 0.5*I*[ω]^[²]/([θ])

A final word of caution here is that my formula above does not allow for efficiency loss, friction or any speed ratio's
which may exist with your set up.

now if you use this torque and transpose the formula for motor power you can find the RPM for the motor ie:-

motor power * torque/(2*[Π])=RPM


I= moment of inertia of cylinder. (haven't allowed for load on cylinder)

[ω]= angular speed in radians/second

[θ]= angular displacement in radians for given second.

regards desertfox
 
"T= 0.5*I*?^²/(?)"

Torque varies with angle?

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I cut/paste the symbols but it didn't work.

I was questioning why the expression for torque would depend on angle theta.

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Hi electricpete

The torque doesn't vary with angle theta, however the workdone or energy would.
At the start of the formula I started with workdone by a constant torque, the further round you take it the energy
requirement increases.
In actual fact I have said theta is a given displacement for a second in which case I could have said [ω] and the formula would have been:-
T= 0.5*I*?^²/[ω]

hope this helps

regards
desertfox

 
Thank you so much for your suggestions. In the end, my calculations gave me a ballpark figure, but I decided to rig something up to actually measure the force required and calculate from there.

Thanks again,

Steve
 
Energy in = energy out.

T1N1 = T2N2/effic

N1/N2 = gear ratio

T1 = motor torque
T2 = output torque

N1 = motor speed
N2 = output speed

This is basic to gearboxes or friction drives.
 
I don't think the problem is really well described:

If the motor was not POWERFUL enough to spin the drum+load then it should have either burned itself up or made the termic relay jump, this has not been stated, what has been said is that it did not achieved to "spin the load" which to me means it did not made it reach enough speed or even move.

If so then the problem is not one of motor torque requiered but one of torque transmision, which probably is caused by the friction transmision from the rubber covered disk to the drum.

I've dealt with that in friction presses and old mechanical variable speed drives and the sistem works fine until it reaches the maximum load transfer that it can transmit and then it starts sliding and loosing grip.

Under that situation one can either increase the contact pressure between disk and drum, change the covering material or surface that are in contact or increase the width of the disk, the problem is that you need a certain amount of sleeping in order to start the motor and accelerate slowly the load but you need to increase the "hold" so that you can accelerate to greater speeds and that calls for some form of pressure variation as the only solution besides a gradual speed increase/decrease with an electronic speed variator.

If the motor did stalled then we have a power problem that can be solved trough your equations.

Besides that, a squirrel cage motor has a set speed that depends on the motor design (2, 4, 6, 8...poles) and the current frequency (50 Hz, 60 Hz or whatever if you have an electronic variator)but at a fixed frequency you will reach a fixed loaded speed which should be 3600/3000, 1800/1500, 1200/1000, 900/750 and so on for 60/50 Hz at 2, 4, 6, 8 poles construction) with an electronic frequency variator you can vary very widely.

 
Yes, the problem was stall/not enough power (not spinning at all). I may have to deal with the friction transmission once the motor is powerful enough; it is not yet a problem.

After measurements and simple calculations, I now know I need 0.75 Ft lb of torque, 60 rpm, and .009 HP (via the equation, HP = torque * rpm / 5252), in an AC motor that is either under 2.5 inches long (and chubby), or less than 1.5 inches in diameter (and lanky).

Any suggestions as to a company good for small high-torque motors would be greatly appreciated.
 
One note on that- if it seems easy to find, it has not been. Most motors that can provide enough power are more than 2.5" long; so far I have only found one acceptable motor, from a company called Crouzet.
 
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