Electric Motor and VSD Relationship 1

[SJ1]

Hello Everyone,

I am hoping someone can clarify some basic issues regarding motor operation for me.

I have a conveyor that has installed on it 1 HP/1750 RPM 3 phase motor with a 40:1 Gear Reducer. I also have installed an inverter to vary the conveyor speed between 5 – 20 RPM. Is it correct to say - that the gear reducer reduces the conveyor speed to 43 RPM and to achieve any speed lower than 43 RPM the motor speed must be reduced with the inverter? Also, does the output HP from the motor vary with speed? If so, how can I determine the output HP at my various conveyor speeds?

I would like to calculate the output HP because during the conveyor operation the motor keeps stalling. I am trying to determine if the system is undersized in torque, HP or both so I can correctly size a new drive system to run the conveyor.

Any input would be much appreciated.

Thanks

 

[TheApprentice]

Here is a great site that talks about AC drives and motors, Hope this helps.

http://www.sea.siemens.com/step/downloads.html

 

[Skogsgurra]

Great link, Apprentice!

But it takes some time to arrive at an answer to SJ1's question wading through all them papers first...

So, Yes. Your motor output is proportional to motor speed. provided that the load torque is constant - which is a fair assumption for a conveyor.

And, yes. Your motor and gear should give you something like 43 RPM when fully loaded. And 45 RPM when not loaded.

A 40:1 gear normally has a rather bad efficiency, especially if it is a worm gear. So do not expect to have much of the torque transmitted to the belt (if it is a worm gear, that is).

And, yes, again. You lower the motor speed using the inverter to achieve speeds between 5 and 20 RPM. But do not expect the motor to output more torque at these lower speeds. It would, if it had been a series wound DC motor, but it is not. It is a normal asynchronous induction motor. And, as such, it has a constant torque (with some negative exceptions) below base speed. So, if it stalls at 43 RPM, it certainly will stall at lower speeds.

For more details, dig into the papers recommended by The Apprentice.

 

[Barry1961]

A standard VSD driving a standard motor can have a 4:1 speed range ratio at a constant torque. This is most often done by VSD varying the hertz from 60hz to 15hz. Through this range you will have a constant torque so your horsepower will change in proportion to speed. At 60hz you will have 36in-lb at 1740rpm which is 1hp and at 30hz you will have 36in-lb at 870rpm which is 1/2hp.

When you run a motor below 15hz the amount of torque you can produce becomes more dependant on the drive and motor combination. With your current setup you are running around 7hz in theroy, but with your load I would guess 10hz with a lot of slip. When you get below 10hz things get fuzzy fast with a standard VSD. Using a worm gear reducer can also hurt at very low speeds.

If you can post the specs of the conveyor such as type of gear box, diameter of head pully/sprocket, conveyor type, total load, product speed and change in elevation it would help.

Barry1961

 

[SJ1]

Greetings,

Thanks to everyone for your input.

Skogsgurra I was hoping you could clarify something for me. You said that I shouldn't expect much torque to be transmitted to my conveyor shaft because of the inefficiency of the worm gear. So, if my full load torque is 36 inch lbs (based on 1 HP/1750 RPM Motor) and I have a 40:1 worm gear - I will not get (40 * 36) inch lbs of torque transmitted to my conveyor? I always thought that the gear reducer also acted as a torque multiplier. So that, the torque actually transmitted to my conveyor would be my full load torque from the motor multiplied by the worm reduction ratio?

Barry1961 thanks for the info on HP and HZ relationship. The specifications for my conveyor are as follows:

- 17" Wide Horizontal Paddle Conveyor (Similar to a screw conveyor but with more mixing) that is 8 feet long.
-Product being conveyed is peanuts (bulk density 37 lbs/ft^3) at a throughput of roughly 660 lbs/h. The conveyor operates a 90% full containing 600 lbs of product load.
-The conveyor needs to operate between speeds of 5-20 RPM.
-The gear reducer is a OHIO Right Angle Worm Reducer with an output shaft of 1 5/8". The output shaft from the reducer is connected directly to the conveyor drive shaft (2 15/16")directly by a coupling.

I hope that gives a little more information.

Thanks

 

[Skogsgurra]

About the torque transmitted.

Yes the gear amplifies the torque. But you lose some because of the internal friction between worm an wheel.

Your conveyor seems to be mostly friction. It also seems to be a little too much (I haven't even tried to calculate the load) for a 1 HP motor. And, as Barry says, it gets worse when you run at lower frequencies. (One of the "negative exceptions" that I mentioned). You can sometimes overcome the stall tendemcies if you increase the boost (increase the V/Hz ratio) at lower speeds.

You need to look out for overheating, though. The cooling is less efficient at these low speeds.

Frankly, I think that you should go for more power. Try a bigger motor and a more powerful drive.

 

[Barry1961]

The efficency of a worm gear is roughly equal to 100 minus the ratio. For example a 40:1 reduction would be 60% efficient, 1hp input gives you .6hp output. The ratings in most worm gear catalogs are generous but will have a disclaimer in the back stating that the gear efficiencys are based on the loading, speed and temperature.

I would recomend buying a 90:1 ratio reducer this would give you the 20rpm output at 1800rpm input which you can get with the VSD. The reducer should be a hypoid or helical bevel type if you need the right angle. This should give you an efficiency of close to 90%. Your low speed would then be around 15hz which would improve your motor performance. This arangement should more than double your available output torque while keeping your current VSD.

I am a little surprised that your are using a worm gear around food. Stober and SEW both make good food grade reducers and are strong competitors. Since you are using a VSD and have a motor already I would go for the Stober since SEW only sells gearmotors and they have had motor problems with VSD's. If you have room for a inline you might be able to get a reducer for $700 with a right angle costing around double that.

If you do decide to try a bigger motor on your worm be sure to check the thermal rating as well as the torque rating.

Barry1961

 

[DickDV]

This is an excellent example of a mismatch between load, power train, and motor.

The system should be designed so that the motor is running at 90Hz when the load is at full speed.

In this case, the ideal situation would be to use a higher ratio gearbox but it sounds like it has already been purchased and installed.

The simplest alternative, in my view, is to exchange the 1hp 1800rpm motor for a 1hp 900rpm motor. The torque will double at all operating speeds. The motor will never develop 1hp because it never reaches base speed but it is far better than using an 1800rpm motor which will never develop even 1/2 hp at full conveyor speed.

You will need to be sure the drive will be big enough to handle the slightly higher current of the 900rpm motor but, even if it doesn't, set the current limit at the drive max limit and run that way.

You will have much more torque and very likely will operate properly.