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#1 bob

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Posted 22 September 2008 - 12:59 PM

Hi,

I am commissioning a VSD driving a conveyor. The VSD is struggling to pick up the load at standstill and is operating in current limit particularly when the conveyor is loaded. I have got a 4 pole motor on the conveyor and my intention is to replace it with a 6 pole one. However, I am not at all convinced that this will work. My feeling is that the motor is not able to break away the load at very low speed. The drive is operating in open loop mode. I understand that the 6 pole motor would provide a better starting torque but my worry is that would it be enough to start the loaded conveyor. I have also tried to con the drive by inputting a bigger motor(200 kW) in the configuration but this did work in some occassions but again with the conveyor fully loaded, the drive ran in current limit mode.The installed motor is a 160 k W one.
Any comment !
Bob


#2 marke

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Posted 22 September 2008 - 07:23 PM

Hello Bob

There are two potential issues here.

The first is the torque required to break the conveyor away, and the second is the torque to accelerate it to full speed.

I usually allow for 180% torque to break away a loaded conveyor, but in some cases it can be higher. (not very often)

One of the problems with a VFD, is that many do not develop a high torque at zero shaft speed. If you use a closed loop vector control, it should be possible to get a high torque at zero shaft speed provided that the algorithm in the drive is correct.
If you use an open loop vector control, you almost certainly will not get the very high torque at zero shaft speed. It is not uncommon for open loop vector to have limited torque below 5 Hz or even 10 Hz despite what the manufacturers say. The result of this, is that the drive accelerates up to a current limit, but because the flux in the iron is low, the slip is high and the conversion of amps into torque is reduced. (a bit like a soft starter with a slip of say 5 Hz)

So, you need to use a VFD that is capable of developing a high torque, (I would suggest 180% FLT) at zero shaft speed, and to develop that torque right through to full speed.
One option is to use a closed loop vector, the other is to use a DTC drive such as the Emotron VFX product. (I have had good success with these under similar conditions where other open loop vector drives have not worked)

Using a lower speed motor and overspeeding it for full speed operation wil certainly increase the low speed torque and is one way to get an improvment provided that the motor is large enough to provide the required power at full conveyor speed. If you overspeed a motor, it becomes power limited, so the torque goes down from it's rated torque.

Best regards,
Mark.

#3 bob

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Posted 23 September 2008 - 11:33 AM

Hello Marke,

You are absolutely right. The drive is very unstable below 5 Hz. However, I would like to discuss further about open loop configuration. In constant torque torque application, a drive is capable to deliver the full load torque throughout the speed range that is from standstill to full load speed. Agree! In my partcular application, very often around 20-30 Hz, the drive is already operating well above full load torque^105-110 % FLT and obviously the current follows the torque and reaches FLC though the available shaft output listed by the drive is below 50 % of the rated capacity of the motor. This is quite paradoxal because although shaft power is still available on the motor , the drive is current limiting because it is approaching FLC and FLT.
Your comment will be much appreciated.

Best regards.

Bob

#4 marke

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Posted 23 September 2008 - 06:54 PM

Hello Bob

When you operate a motor on a VFD, the motor can only operate at a maximum flux density. As the flux determines the maximum torque that is available from the motor, we describe the operation of the motor below rated frequency as being torque limited and above the rated frequency, the voltage is limited so that as we increase the frequency, the flux reduces. This creates a power limited situation.
The shaft power of a motor, is equal to the shaft torque times the speed, so if the motor is torque limited below full speed, the effective power rating of the motor reduces with speed. If you have a 110KW motor at rated speed, it can only deliver 55KW at half speed. I thik that this is what you are seeing. At half speed, and rated torque, the motor will draw rated current (FLC) but the terminal voltage is half full voltage and the KW is half rated KW.

Best regards,
Mark.

#5 r_mac

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Posted 09 October 2008 - 04:31 PM

These are all good points but I have not read what type of motor was being used. An inverter rated motor versus an off the shelf standard motor. The selection of the motor for the application is equally important. I would suggest looking at the published torque curve of the motor installed versus an inverter duty rated motor if one was not initially used. That may get you over the hump.

Best of luck

Robert




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