Starts/minute

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ptonsparky

ptonsparky

Tom
Location
NE (9.06 miles @5.9 Degrees from Winged Horses)
Occupation
EC - retired
We inherited a small industrial that produces hay pellets. Last year we installed PLCs with touch screens for the mostly manual control of motors. After the second burnout of one motor we discover that the operator is purposely jogging a group of motors to ?fix? the problem of to much raw product on a conveyor. I will use the PLC to limit the number of starts for this group of motors but am curious as to how many starts per minute may be acceptable. The motor is 3hp@480v and draws 2.8 amps under normal operation. It is a high inertia load via a gear reduction. I have no idea of how many times he may have reset the overloads before burnout.
 
Jogging duty is often listed as 5 openings and closings per minute and 10 total in 10 minutes for small horsepower starters.
 
The data I have seen from motor manufacture's is 6 starts per hour. It may be best to contact the motor manufacturer with the model number and see if this is a special duty motor and what the limitations are.

Tony
 
Assuming (based on the 480V spec) that this is a NEMA design motor, the Starts-per-hour is not a hard number. It is a formula from columns in a chart (NEMA MG10 Table 2-3). The NEMA formula is;

Maximum Starts/Hour = The lesser of column A, or column B/Load WK^2, With column C between start attempts where;

A = Direct Maximum Starts/Hour column, based on # of poles,
B = Load multiplying factor column from the same chart
C = Minimum rest time in seconds between start attempts.

So for example, if your motor is 4 pole (1750RPM), and your load inertia (WK^2) is 1.5 lb-ft^2, then column A = 19.8 S/H and column B = 17, so 17/1.5 = 11.33, then you would need to use the 11.33 value. If however your load inertia is only .5 lb-ft^2, then B becomes 17/.5 or 34, so then A is lower and you can use the 19.8 S/H rating. In either case, C on that motor is 40 seconds, so you must have 40 seconds between starts for this to be valid. If you exceed any of these values, all bets are off unless you speak to the motor manufacturer.

PS
Ah, you posted as I was writing. Good to see that you checked with the mfr., that trumps everything.
 
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The mfg did indicate that less time between starts would lesson the Starts per hour. I did manage to download that table. Determining factor B would be a shot in the dark without removing the drive, for me at least. There is another larger motor that starts at the same time as this so that will factor in as well. The manager has decided an audible alarm and text display of impending Job Loss will be enough for now.

Thanks again.
 
Sounds to me like a VFD may be in order so the operator can slow down the flow of raw material without stopping a motor.
 
Raw product is loaded on to the conveyor with a large front end loader. The operator is supposed to make sure the product is evenly distributed. The conveyor is on a vfd and travels at inches per minute. The equipment is overwhelmed when +4' instead of -2' is dumped on.
 
You can overcome this issue by putting in a larger motor. More iron = higher tolerance to rapid cycling. I'd toss in a 5HP motor (if you can make the larger frame size work for you), then use a soft starter on it with just a 2 or 3 second ramp to keep the extra torque from tearing up the drive train. The soft starter will acto to dampen the operators overshoot a little as well, might make it so he doesn't need to repeat the bumps as often.
 
It is just plain operator error. They know the limitations of the equipment, but in their defense the hay does not always exit the loader bucket as planned. The correct procedure is to manually remove excess raw product fom the conveyor. A larger motor in this case would just move the problem to the next step.
 
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