Eng-Tips is the largest engineering community on the Internet

Intelligent Work Forums for Engineering Professionals

  • Congratulations waross on being selected by the Tek-Tips community for having the most helpful posts in the forums last week. Way to Go!

Centrifuge AC Induction Motor Requireing 30 Starts Per Hour

Status
Not open for further replies.

rglassburn

Mechanical
Oct 18, 2007
11
0
0
US
Hello All,

I have been asked to specify a motor for a new centrifuge. It needs to be able to handle 9.4 lb-ft^2 of inertia and spin at 3425 rpm. In order to cycle the number of parts that I need, I will need to be capable of cycling the centrifuge from 0 to 3425 rpm up to 30 times per hour. The centrifuge will be driven by a VFD. My plan is to direct drive the table using a flexible coupling.

As I have been studying the requirements, I have found the MG 1-40.40 table that seems to indicate 10 HP motor. Since a VFD will be needed to control the speeds, I will pick an inverter duty motor. However, looking around a bit more, I found MG 10-2001 which has table 7 "Allowable number of starts and minimum time between starts for Design A and B motors" which indicates that a 10HP 2 pole motor would not be able to start that many times per hour. It shows a little more than 1 start per hour?

Can anyone provide some clarity and/or point me to a source that can help me with this spec?


Thank you in advance for your insight!

Thanks,
Rob

 
Replies continue below

Recommended for you

From the data presented, I would use a two-pole 5hp TENV motor with a regen drive rated at 7.5hp heavy-duty (or constant torque).

A TENV cools itself about the same whether it is running or not which would be an advantage during the idle time in the application.

You will minimize any overheating issues in the drive by sizing it the way I suggest. A regen drive would be the right choice because of the amount of energy being returned back to the line during braking.

Suggestion: use a Marathon Black Max motor. They may not offer a two pole version. In that case choose a four pole 5hp and run it to 120hz. The motor is guaranteed to be constant hp up to that speed.

Choose a drive with good sensorless vector capability. Using flux vector with an encoder is expensive overkill.
 
I've worked with the Marathon Black Max motors before some years ago (just small ones, though... 2HP or so). They are remarkably heavy for their size, and just give you a feeling of 'trustworthiness'. I've never had a problem with them, and they certainly worked fine with inverters... I've run them at over 5,000rpm for long periods with no problems. I can only assume that they are as good now as they were then.

Mike
 
A TENV cools itself about the same whether it is running or not which would be an advantage during the idle time in the application
I would be curious to understand the basis for that statement. Even TENV still have rotor features to promote internal fan action which serves to cool areas such as the endwindings. It does not sound logical that the OEM's would sacrifice efficiency for fan action if it doesn't have an important function.

=====================================
Eng-tips forums: The best place on the web for engineering discussions.
 
I should mention that I'm going by memory. I have some photo's of a TENV rotor at work that I'll check on Monday

=====================================
Eng-tips forums: The best place on the web for engineering discussions.
 
ePete,

My understanding is similar to yours - a TENV motor has an internal 'fan' on the rotor to promote air movement within the casing. I don't see many TENV machines but the handful of DC ones I have seen recently enough to remember definitely had this feature.


----------------------------------
image.php

If we learn from our mistakes I'm getting a great education!
 
Thank you all!

Very useful and helpful information.

The flexible coupling is indeed like a "lovejoy" flexible coupling. I will make sure that the drive and load are well aligned even without the flex coupling - as well.

I have chosen a 5HP TEBC(Totally Enclosed Blower Cooled) motor. It is a monster weighing in at 120lbs.

I will start looking into the economics of regen vs using shunt resistors.


Thank you,
Rob

 
The table is connected to the machine frame thru 2 bearings located on a shaft connected to the table. The motor is vertically mounted and connected to the frame. The frame will be welded together. I am allowing adjustment room in the bearings and the motor for alignment.

I am going to lock down any fasteners with lock washers or safety wire to keep them from vibrating loose. This should allow the alignment to stay constant.


Thanks,
Rob

 
Just to address an earlier statement so that you don't go off half-cocked...

Your motor supplier told you that their motor could go to 6000RPM. Don't misconstrue this. Yes, the VFD can supply a higher frequency and Yes, apparently that motor's bearings and cooling system allow for the higher RPM. But in case you are not aware, as you increase a motor's speed with a VFD beyond its base rating, i.e. 4 pole = 1800RPM, you start to operate in a "constant HP" mode, meaning that your torque is decreasing at a non-linear rate. So taking an 1800RPM motor to 6000RPM is theoretically possible, as long as your load torque requirements diminish at a similar rate.
 
Jraef,

That makes sense to me. In order to accelerate my load up in the time that I need, I need a 1.7 lb-ft of torque on average. The motors full load torque rating is 15.0 lb-ft of torque.

Which of course seems like way too much for the need. Unless this diminishing HP with increasing RPM is taken into account.

Thank you for the comment.

Rob
 
I like to use the European style motors for frequent starting and stopping applications. The smaller diameter (compared to the NEMA size) will cut down the motor inertia, a good thing when starting and stopping. SEW-Eurodrive as a reference.
 
rglassburn,

The 5 or 10 HP motors were making some sense to me. Assuming the 9.4 lb.-ft^2 units were Force-Length units, the mass inerta is about 0.3 slug-ft^2. With,say, 2 lb-ft of torque, I calculate over 500 seconds to reach 3400 RPM. This is why I asked what the Inertia units really are.
 
sreid,

The units are lbm-ft^2. So they are pound mass units [no-one likes slugs anymore :-( ] Typically called WR^2 or WK^2. Simply the weight times the radius of gyration squared.

The time in seconds to change speed is given as follows:
t = (WR^2)*(change in RPM speed)/[308*(avg. accel torque)]


Thanks,
Rob
 
rglassburn

Clipped from one internet site

POLAR MASS MOMENT OF INERTIA
The polar mass moment of inertia (J) is the measure of resistance to a change in rotational velocity about an intended rotational axis. It may be derived by dividing the flywheel effect WR2 by the acceleration due to gravity.

J = WR2/g lb-ft-sec2 where

R = Radius of Gyration, ft
W = Weight, lb
g = 32.2 ft/sec2

So is your inertia 9.4 lbm-ft^2 ?
 
Hmmm...

I calculate:

9.4 lbm-ft^2 = 0.29 lbf-ft-s^2 (=slug-ft^2)

dividing by "g" (32.2 ft/s^2) as sried does.

For acceleration time, I calculate:

0.29 lbf-ft-s^2

* 3400 rev/min * 2Pi/rev * min/60s

* 1/(1.7 lbf-ft)

= 61 seconds acceleration time

So in a 2-minute cycle, you wouldn't quite make top speed at all. Of course, with your motor capable of more torque, you would be able to accelerate and decelerate faster than this, but I suspect you will want substantially higher torque levels.

Curt Wilson
Delta Tau Data Systems
 
Status
Not open for further replies.
Back
Top