Soft Start Problem

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#1 gokulkrish2

gokulkrish2

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Posted 02 July 2008 - 04:51 PM

Hello All,

Im a new member to this forum. I am a Graduate student and im doing an internship in a Mining Company. I am working on a conveyor project whose problem is that it wont start when it is shut down loaded. It is a horizontal conveyor and when i analyzed the data about Load Current at different instances, the maximum current for a maximum load was only 180 amps whereas the motor is rated at 270 amps. So i figured out it is a prob with the starting current or the starting torque.

I know that the conveyor has 250HP motor which is connected to the pulley through a fluid coupling. The service factor of the motor is 1.15 and the full load current of the motor is 270 Amps

I need to know whether we can take off the soft start and start the motor across the line so as to get the full torque?

Im also concerned about the protection of the transformer which feed the motor. (inrush??) The transformer is connected to the mains through a 50amp fuse on the primary side.

Can someone help me on this issue. Thanks in advance.

regards,
gokul

#2 marke

marke

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Posted 02 July 2008 - 07:27 PM

Hello Gokul

Welcome to the forum.

You are correct, the motor is not producing sufficient torque to start the loaded conveyor. - From my experience, this can be over 150% torque for some materials conveyors.

You can increase the starting torque, by increasing the start current.
Have a look at http://www.lmphotoni...m/m_control.htm and http://www.lmphotonics.com/m_start.htm which will give you some more background on this subject.
Basically, as you reduce the start voltage, you reduce the start current. (ohms law) As you reduce the start current, you reduce the power in the rotor by the start current reduction squared. (start torque)

You need to increase the start current/voltage to increase the torque and the loaded conveyor will start provided that the motor develops enough torque at full voltage.

In you case, you also have a fluid coupling between the motor and the conveyor drive. The objective of the fluid coupling is to reduce the start torque snatch and reduce the maintenance required. In order to cater for an unloaded conveyor and a loaded conveyor, the fluid coupling should be a delay fill type. This will transfer a low torque initially and increase the torque transfer over time.
With a delay fill fluid coupling, the motor will start easliy as it only sees the torque transfer of the coupling, and it will reach full speed at a low start current ( 300 - 400%) and the conveyor will start after the motor is running at full speed.
With a standard fluid coupling, the motor must produce more than the torque transfer of the fluid coupling during start, in order to reach full speed and allow the coupling to control the conveyor start. If the start torque produced by the motor is less than the transfer torque of the fluid coupling, it is achieveing nothing and is adding another inefficiency to the system.

Best regards,
Mark.

#3 gokulkrish2

gokulkrish2

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Posted 02 July 2008 - 09:02 PM

QUOTE (marke @ Jul 2 2008, 12:27 PM) <{POST_SNAPBACK}>
Hello Gokul

Welcome to the forum.

You are correct, the motor is not producing sufficient torque to start the loaded conveyor. - From my experience, this can be over 150% torque for some materials conveyors.

You can increase the starting torque, by increasing the start current.
Have a look at http://www.lmphotoni...m/m_control.htm and http://www.lmphotonics.com/m_start.htm which will give you some more background on this subject.
Basically, as you reduce the start voltage, you reduce the start current. (ohms law) As you reduce the start current, you reduce the power in the rotor by the start current reduction squared. (start torque)

You need to increase the start current/voltage to increase the torque and the loaded conveyor will start provided that the motor develops enough torque at full voltage.

In you case, you also have a fluid coupling between the motor and the conveyor drive. The objective of the fluid coupling is to reduce the start torque snatch and reduce the maintenance required. In order to cater for an unloaded conveyor and a loaded conveyor, the fluid coupling should be a delay fill type. This will transfer a low torque initially and increase the torque transfer over time.
With a delay fill fluid coupling, the motor will start easliy as it only sees the torque transfer of the coupling, and it will reach full speed at a low start current ( 300 - 400%) and the conveyor will start after the motor is running at full speed.
With a standard fluid coupling, the motor must produce more than the torque transfer of the fluid coupling during start, in order to reach full speed and allow the coupling to control the conveyor start. If the start torque produced by the motor is less than the transfer torque of the fluid coupling, it is achieveing nothing and is adding another inefficiency to the system.

Best regards,
Mark.

Thank you for your speedy response marke. Actually i got the information that we are using a delay fill fluid coupling. Actually, now im not clear on whether i should do the changes on the motor side like to do the changes on the soft start to increase the voltage across the terminals or on the fluid coupling to decrease the starting time of the fluid coupling.

Can you suggest me wat to do?

And i need one more suggestion.
Can i take out the soft start and just start the motor across the line?

Thanks
gokul

#4 marke

marke

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Posted 03 July 2008 - 07:16 PM

Hello Gokul

I would suggest that the best solution is to lift the start current of the soft starter so that it develops enough torque to quickly spin the motor up to full speed before the fluid coupling increases the torque transfer. This will give you the advantage of a reduced starting current.
Then, the motor, operating at full speed can deliver up to it's maximum torque which may be in the order of 200% torque for a relatively low current of 200 - 250% current. You will not achieve this torque with the motor operating at less than full speed except by operating with a slip ring motor.
Provided that the maximum torque transfer of the delay fill fluid coupling is higher than that required to break away the conveyor, it will start. If the maximum torque transfer of the fluid coupling is less than the required breakaway torque of the loaded conveyor, you will not be able to start it.

If the start current of the soft starter is too low, the motor will not get to full speed and so you will not get maximum torque transfer trough the coupling.

step 1. Increase the start current so that the motor can easilyy spin up to full speed even though the conveyor may not be moving. This may require a start current of around 400%. If the soft starter is a voltage ramp type, I would suggest that you lift the initial start voltage to around 60%.
Step 2. check to se if the maximum torque transfer of the coupling is sufficient to start the loaded conveyor. If you monitor the current drawn by the motor once it is at full speed, you should see the current rise as the torque transfer increases. You should also be able to determine when the coupling is "full" as the current will not rise any further. If the current reaches a maximum and it is less than say 200%, then the coupling is limiting the torque and an adjustment to the coupling may result in a higher torque transfer, or an alternative coupling may be required. From the current, you can get an indication of the torque being transferred and make decisions from there.

Replacing the soft starter with a DOL will not achive any additional torque provided that the soft starter is set up to accelerate the motor to full speed before the load is transferred by the coupling.
Removing the coupling is going to reduce the maximum torque transfer and is OK if the starting torque is less than 100%. If the breakaway torque is in the order of 180%, then you will have difficulty achieveng this with a soft starter and no fluid coupling.

Best regards,
Mark.

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