AC inverter rated motor

[lan3]

My question is: I have a centrifuge that will spin a 10 inch diameter drum that weighs 20 lbs at 6000 rpm it can take up to 10 seconds to get up to speed..What HP motor do I select?? What is the equation for this??

 

[RadarRay]

10"=.8333Ft diam = .4166ft radius
WK^2 = weight*(r^2/2 + r^2/2) for a hollow cylinder. Not quite true here but should be a good estimate.
wk2=20*(.4166)^2 = 3.47#ft^2

Tq to accel = (wk2 * delta rpm)/(308 * time in sec)
Tq= (3.47*6000)/(308*10) = 6.77 #ft

HP = Tq*base speed/5250 = 6.77 * 6000/5250 = 7.74HP

This assumes no losses and that the 6000rpm is the base speed of the motor. Especially note it did not include the HP required to accel the motor itself. So if you found a 10HP, 6000rpm motor, it should work.

Note if you go to a lower base speed motor and gear up, you still will require 7.74 for the centrifuge, you still have to include the motor and you will also have to include any gearing. By choosing 6000 rpm, I was saying it was direct geared, although this is not likely.

 

[lan3]

Thanks for the help...

 

[RadarRay]

I forgot something. Since a motor tied to a drive is typically capable of an overload, you can use the overload capability to reduce the motor size. Typically a constant torque set up would be good for 150% which would mean you could probably use a 7.5HP motor, maybe even a 5HP if you can get 200% out of the system for 10 to 20 seconds.

 

[lan3]

Is there a simple conversion between torque in lb-ins to hp or HP to lb-ins??

Thanks for your help

 

[jbartos]

Suggestion: Conversion:
T[lb-ins]x[ft/(12xin)]=w[rad/sec]xP[Watts]x[HP/(746xWatts)]