induction motor speed variations

[JIMGEN]

We have two new identical 350hp, 50hz, 1500rpm , 380v motors (derated from 400hp, 460, 60hz)

running on sheaves and belt driven common shaft. We are showing a 2 rpm difference between the two motors. (And significant load differences)

How much, if any, rpm differential is allowed by NEMA or IEC? I am not sure if the motors have sequential serial numbers but they were ordered on the same document.

TIA for reply and comments.

 

[jraef]

With an induction motor if the loads are different, the speeds will be different. More slip. The "RPM" rating on the nameplate is the RPM at rated load. Anything more or less will result in a slightly different speed.




"You measure the size of the accomplishment by the obstacles you had to overcome to reach your goals" -- Booker T. Washington

 

[waross]

A worn set of belts will sit deeper in the sheaves. This will throw the drive ratio off slightly. It could easily account for a 2 RPM difference in speed.
If the motors are rated for 1760 RPM then the slip speed is 40 RPM. The difference between no load and full load is less than 40 RPM.
If the motors are less than fully loaded and running around 25 RPM slip you could have 8% or more difference in current draw.
The other possibility is that one set of belts is slipping and that motor is running a little fast and pulling less than its share of the load. On a multi-belt drive, mis-alignment may throw the load onto one or two belts out of the set. Those belts will slip and suffer accelerated wear.
Check the belts for wear, tension and alignment.


Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[7anoter4]

I agree with waross, of course.
In my opinion the OP presents the following-simplified-case:
Two-induction [squirrel cage] - motors together act a loaded shaft. The sum of both torques [or powers-as the same] equals the load torque.
If one of these torques will be less than a half of load torque the other motor has to increase its power in order to equal the sum with the load torque. So if the total efficiency -motor + transmission-gearbox, sheave, belt-will be less in one side and the power reaching the shaft will be less, then, the second motor has to increase its power.
If the motor rated velocity is about 1480 rpm then 20 rpm is the slip for 350 hp.
2 rpm it is then 10% more slip or 10% more torque or output power [approx.].
That means [for instance] if one motor is loaded 335 hp the other will be 365 hp.

 

[mikekilroy]

Jimgen, if you come back, please tell what "And significant load differences" really means in some sort of engineering units; % hp, & torque, % amps, etc. Without this detail, it is simply guessing what you have and thus no clue if it is a REAL problem or not.

As 7anoter4 showed, 2rpm diff should be only about 10% torque difference; hardly worth worrying about?

As waross stated, pulleys and belts can wear and allow one motor driving the same mechanical load to go slightly different speed. Some math for a sanity check: You did not say what the transmission is from motor shaft to load so let's assume 12" pulleys somewhere... So C=37.70" If one belt sits 1/16" lower in the pulley, it only goes around 37.50", a 0.5% difference, so I would expect one motor to go 1480rpm and the other 7 rpm slower. You only see 2 rpm difference - I question if this is a problem or worth worrying about? Your process needs to be able to handle these slight differences.

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