Calculating Horsepower required to spin a cylinder 1

[azonicbruce]

Let me apologize in advance for not knowing the basics; it's been awhile since I've needed to use them.

I need to determine what the HP rating needs to be for a motor that will turn a Drum Magnet used for ferrous separation systems. Here's and example of what it looks like:

http://www.sgm-magnetics.com/pictures/19-3a2.jpg

This is what I do know:
Weight of Drum: 39,860 lbs
Drum diameter: 60"
Operating speed: 24rpm

I've tried the following approaches, with the following assumptions:
Calculate Work, then convert into HP based off of RPM.
Since scrap is only on the drum for approximately half the circumference, the distance work is being done over is half that. Since I'm not sure exactly how much steel scrap is on the drum at one time (probably 200-400 lbs), I'm assuming that at least the weight of the drum is being worked on.
So,
39680lbs x 7.85ft(half the circumference)= 311,488 ft-lbs of work.

1HP=33,000ft-lb/min
So if the drum is spinning 24rpm, then it is covering the 7.85ft of distance in .021 min.

This comes out to 449HP. Right now most customers are using 20HP motors; we're trying to see if we can get by with less, so my number is way off.

I tried this way too:
HP=(TorquexRPM)/5252
Torque would be 39,680 lbs 2.5ft(radius of drum)=99,200 ft-lbs

This comes out to 453 HP.

I know I'm doing something wrong, I'm just not sure what. Either I'm using the weight of the drum incorrectly, or using and incorrect distance over which work is performed, or improperly using the time over which work is performes, or I'm just plain mess up and going about it all wrong.

Any thoughts?

 

[azonicbruce]

Just doing a little thinking...

Should I not even be looking at the weight of the drum and focusing more on the amount of scrap metal that the drum needs to dump off? At almost 20 tons and 24 rpm I'd think that once the rotor got going, it'd have quite a bit of inertia and would therefore need the power to keep moving the scrap that keeps getting thrown on it.

If that were the case, then 400lbs of scrap covering half the drum, would mean that 800lbs of scrap is being moved each revolution.
800lbs x 15.71ft(circum.)=12,568ft-lbs
12,568ft-lbs x 24rpm = 301,632 ft-lbs/min
301,632 ft-lbs/min divided by 33,000 ft-lbs = 9.14 HP

Am I still on the wrong track?

 

[sreid]

Assume that the 400 lb of steel is always at the worst point on the drum. This is the entrance point in the horizontal plane (when the load is at the top, it produces no gravity torque and when the load is on the down moving side of the drum it helps move the drum).

The torque is then the weight times the drum radius (2.5 ft) so

T = 400 x 2.5 = 1000 lb-ft

The drum RPM is 24 RPM so the drum angular velocity (Omega) is

Omega = RPM/60 x 6.28 = 2.512 Radians/sec

Power is then

Power = T x Omega = 1000 x 2.512 = 2512 lb-ft/sec

Horse Power is 550 lb-ft/sec so

HP = 2512/550 = 4.57 HP

 

[EnglishMuffin]

To calculate the torque needed, one would have to understand the way in which the separator operates. The work done per rev would presumably be related to the force needed (if any) to remove the material from the drum and any frictional losses that might be present, as well as the fraction of the circumference occupied by the material at any given time, together with the angular location of that fraction, but it has very little to do with the weight of the drum, assuming it is symmetrically disposed. The moment of inertia of the drum (and by implication its weight) would only be important from the point of view of acceleration time. Could you perhaps clarify how the separator works, giving some indication of the forces to be overcome?

 

[sreid]

Since this is a magnetic separation process, there is probably a fair amount of torque required to move the drum past the (Electro?) magnet due either reluctance or eddy current forces (or both).

 

[MintJulep]

Why not just measure the current draw of a motor on an existing unit in operation?

 

[Barry1961]

One common HP equation is (torque(in-lb) * RPM)/63025

There is going to be some friction in the bearings. Also you will probably want a slow start device to reduce the torque required for acceleration.
It also looks like there may be some torque resulting from material "pouring" down surface of drum near the left quadrant. This could cause a lot of torque depending on the surface of the drum.
I agree with sreid that load torque needs to be calcualted at worst case.

Some of these factors would be very hard to calculate without a lot of testing and sampling of materials, so you might want to go with what you know works.

Barry1961

 

[pmover]

motor needs to be sized to overcome drum breakaway torque and bring to to operating speed. how long to achieve operating speed is another design consideration. once at operating speed without material handling, motor hp will likely not be as great as at start-up. however, motor sizing will need to consider all operating scanerios during the material handling operations. without knowing details, note that material magnetized to drum is on upward rotation of drum (need work/power) and when release from drum, the effect gravity is a consideration.
have you considered contacting mfg of magnetized drum and obtain recommendations from them?

good luck!
-pmover

 

[azonicbruce]

Yes, we are in the process of contacting the manufacturer. Also, I have been thinking about startup time. I think that's how I got the 449HP. That would be approximately what it would take to get the 20ton drum up to 24rpm in half a revolution. I'll have to investigate whether or not starup time is an issue or not.

Thanks for all the suggestions, they should help. I know that there are tons of other factors that determine what the exact requirement will be, I was mainly wondering though if I was going about my calculations correctly assuming other things are negligible.

 

[ccw]

Hi Azonicbruce,
You are correct to discount the rotor weight. From an energy standpoint there is no work being done on the rotor, ie. no potential or kinetic energy change while at speed (24RPM). There is some rotor bearing and speed reduction bearing/gearing friction and windage to be overcome but that is probably only a few HP.
The torque required to have the rotor comb through the introduced trash and the starting torque transient are the only significant power demands on the drive motor. Start up transient power demands can exceed the motor placard continuous power ratings by 50 to 100% without harm, provided the motor is not being started in repeated quick succession. Reduced voltage soft start systems can be used if the start up transient demands are too great for the commercial mains or drive train to support.
See NEMA MG-1 spec. for more help on proper sizing of motors.
Finally, MintJuliep has a great idea. For AC induction motors, the motor current, over a considerable slip speed (say 80 to 99%), is analagous to torque.