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Starting an overland conveyor


marke

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I am facing a starting problem for 1.5 km conveyor which drived by 3 x

480 kV 3.3 kV Siemens Motor. The starting method is using 8 steps

rotor resistor. Since the stall problem is alarmed by thermal O/L

protection relay, I want to investigate wheteher the locked rotor is

large enough to handle the static load. Refer to

http://www.overlandconveyor.com/pdf/drive_...ms_sme_2003.pdf

page 5, I want to find the NEMA nameplate of Siemens drive, but I can't

find it in its name plate. I just found unique identity ROTOR : S in

its name plate exept other common data (kWh voltage etc). What is the

meaning of this data?

 

Thank you for the helps.

 

regards,

Wahyu pamungkas

This is a copy of a question posted on my Yahoo forum and well worth discussing here.

 

Best regards,

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Hello Wahyu

 

A materials conveyor started fully loaded can require 160 - 180% torque to start it. There are various ways of starting this conveyor as illustrated in the paper that you referred to.

 

In order to start this type of conveyor with load, you need to develop sufficient torque to cuase the driven load to break away. This is a problem with many starting means, particularly with large squirell cage motors where the Locked rotor torque in insufficient to create breakaway.

On method, not covered in that paper, is to use a slip ring motor.

 

The slip ring motor is a motor where resistors are inserted into the rotor circuit, altering the torque speed curve characterisitics of the motor. Increasing the effective rotor resitance, increases the slip at which maximum torque is developed. If you have a true zero resistance rotor, then maximum torque would be at zero slip, (synchronous speed) and nor torque would be developed below full speed.

The rotor resistance can be selecte such that the maximum torque is developed at 100% slip, or at zero speed. This would ensure that there is sufficient torque to break away the conveyor, but there would be a high step in torque at initiation. The step would be from zero torque to around 250% torque and this may result in mechanical damage and would certainly be far from ideal with an unloaded conveyor.

It is common for unloaded materials conveyors to require a start torque in the order of 40% - 60%, so I would suggest that the initial rotor resistors are selected for a maximum torque slip of around 200% to give an initial torque of around 50%. The second step could be at around 150% slip, the third at around 120% slip etc to give a ramping torque. You would need to have a number of stages at less than 100% slip as well in order to prevent high current and torque steps at resistance changes.

If you keep the conveyor operation at a slip less than the maximum torque slip, you will have lees chance of applying maximum torque to the conveyor and thereby minimise the potential damage to the conveyor.

 

Where a number of motors are being used to drive the conveyor, it is important to ensure that the full load slip of the motors is identical, otherwise, the load will not share and the motor of lowest slip will do all the work. With multiple wound rotor motors, it is common to leave one stage of resistors in the rotor circuit to ensure that the slip characteristics of each motor are the same.

 

In answer to your question, you will not find the Locked Rotor Torque for a would rotor motor as this is a function of the resitance connected to the rotor. You have control over this!

 

Best regards,

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  • 2 weeks later...

Marke,

Thank you for the answers. Now, it is clear that I should check the rotor resistance value. but I am confused with your terminology 150% slip, 120% slip and so on. Is it same as 150% torque and so on?

 

I am also advised that I should consider the steping time interval (I use 8 steps and it is 2 seconds interval between steps). I can't find any equations between slip vs time. I understand that in-accuration justification could lead to motor stall problem. Could you advise how to determine this time manner?

 

I've got infromation from manufacture that our drive is able to give output 2.4 Breakdown Torque at rated voltage. It means it is more than enough to accerate our conveyor in 60 seconds. They suggest me to consider time interval of steping rotor resistor for starting bigger load. It looks like that I face a problem of stepping time rather than capacity problem.

 

Thank you so much.

 

regards,

 

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  • 1 month later...

Hello Wahyu

 

First of all, sorry for the very slow response to this question. Your post came in while I was overseas and it slipped under the radar.

 

When you design secondary resistors, it is normal to select the value of the resistor for the speed of peak torque. Slip is a way of expressing speed, but is the actual figure that determines the rotor chartacteristics.

By definition, the slip of the motor, is the difference between the running speed of the motor and the synchronous speed. A slip of 0% would have the motor operating at synchronous speed, and a slip of 100% would have the rotor stationary. A slip of 150% has the rotor spinning at half speed in the reverse direction to the synchronous speed.

The speed of maximum torque for a given rotor resistor is the speed where the rotor resistance is equal to the reactance of the rotor. As the slip changes, the frequency of the rotor voltage alters. At 0% slip, the rotor current frequency is zero Hz. At 100% slip, the rotor frequency equals the line frequency and at 150% slip, the rotor frequency is equal to one and a half times the line frequency.

 

The step time is dependent on the required acceleration rate and the load inertia. You could use the stator current as a means of changing the rotor resistance. i.e. step when the line current is less than rated current.

 

Best regards,

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