Time-delay between starting of multiple motors 3

[RalphChristie]

I have a question regarding the starting delay between two or three motors driving the same load. In my case it is three motors driving a conveyor. It is 3.3kV, 500kW slipring motors (WRIM) and each one is started with a vapourmatic(liquid) starter with three steps. On the third step the rotor is short-circuited.

If the three motors are started simultaneously, there will be a very high inrush peak.
If the motors start one after another, and the delay between each starting motor is too long, the first motor will do most of the work during starting, while the last motor will do much less. (especially when you want to start a fully loaded conveyor.)

(The question also refers to the delays between each starting step of the motors)

Thus, something like:
Start motor 1
time delay
start motor 2
time delay
Start motor 3


Motor 1 step 1
time delay
Motor 2 step 1
time delay
motor 3 step 1

etc.

In my opinion the ideal time-delay lays somewhere between imediately and say three seconds. Does 500msec sound fine? Too long? Too short? I would suspect the time delay to be the same for starting and between steps.

Many thanks
Regards
Ralph



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[Laplacian]

This is just my opinion, but here goes:

Start Motor 1,
Time Delay equivalent to acceleration to full speed of Motor 1,
Start Motor 2,
Time Delay equivalent to acceleration to full speed of Motor 2,
Start Motor 3,
Time Delay equivalent to acceleration to full speed of Motor 3,

Motor 1 Step 1 = Motor 2 Step 2 = Motor 3 Step 3, etc.

 

[Laplacian]

EDIT -

Motor 1 Step 1 = Motor 2 Step 1 = Motor 3 Step 1, etc.

Sorry.

 

[Barry1961]

I would lean toward the .5 seconds. If you need 3 motors to run this conveyor I would think they need to start near the same time.

Can you stagger the delay on the first step of the starters? You could stagger the start and reverse stagger the first step so the motors all did step 2 & 3 at same time.

I am guessing that your conveyor bed is rollers so the starting friction should be near running friction.

You may be able to use a fluid coupling with an adjustable fill rate to let the motors start unloaded. You could then stagger the delay on the fluid coupling in the reverse of the stagger on the starters. This should also allow a longer stagger on the starters.

Barry1961

 

[jraef]

Interesting problem, made more so by the WRIMs and yet again by the liquid resistance starter. Certainly there is an imersion time to deal with on the LR starter probes. I doubt that 500msec would even be noticed in that light.

I assume the conveyor is unloaded at start and the motors coupled because if they had clutches this would not be an issue. If so, the first motor is going to have to drag the WK² of the other two as additional load, which would increase it's acceleration time over the others. Is there no way to slowly ramp the motors up with the LR starter instead of using specific steps? That is usually one of the main advantages of LR over resistor grids. That way you could limit the starting current and have the motors share the load equally (or close to it).

If by inrush, you meant true magnetizing current inrush as opposed to starting current, then a slower imersion of the probes should help that as well.

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[RalphChristie]

Let me first explain the vapourmatic principle:

The starter consists out of some tanks filled with a liquid. (electrolyte) In each tank is probes that hang (always) in the electrolyte. The number of tanks depends on the size of the motor. All these probes/tanks (for each phase) are in parallel, and everytime a next step came in, more probes/tanks are switched into the circuit. (also in parallel)

When a electric current is passed through the elctrolyte two actions take place:
The electrolyte heats up and its resistance decreases.
The electrolyte vaporises and its resistance increases (±50 times the value of the cold liquid electrolite)
Heavy rotor currents flow at the instant of the start. These currents cause an immediate formation of vapour bubbles on the elctrode blades (±10ms) increasing the resistance and decreasing the current proporsionally. This decreased current cannot maintain the vapour condition and the vapour recombines with the liquid and the resistance decreases to a low level giving a smooth accelerating torque of the motor.

My problem is to do the upstream protection (transformer and incoming feeders) and I am thus looking for the most effective starting sequence. Under normal conditions the conveyor is always started unloaded and no problems are experienced, but when started fully loaded for one or another reason (pull rope, stop pressed, etc) - upstream trips occur. Now to increase the upstream protection (time or pick-up) would be no problem, but then the protection of the upstream equipment would be less sensitive. (especially during faults) Thus, before doing any upstream changes I want to ensure everything downstream are working on the most effective way possible.

Although fluid couplings or something similar would be a effective sollution, it is not possible on this moment due to cost issues and major changes on the drive constructions. Also keep in mind that we have not just one conveyor, but eight with similar drives. The same would apply to changes on the motor starters. (Enough money for technical projects/improvements are always an issue)

On this moment I feel the best thing to do would be to do some tests on the plant. Stop a fully loaded conveyor and play a bit with the starting sequences. I have not though about the stagger and reverse stagger idea, but I will look into it. And like always, production will also be an issue....

Thanks for the comments
Regards
Ralph



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[fsmyth]

Interesting problem.

What is the normal starter timing (conveyor unloaded)?
Seems like about 1/2 that would make a reasonable
first approximation for the motor sequence period (i.e.,
the period from Motor1 Step1 to Motor2 Step1, etc.).

How much longer is the start step sequence when
fully loaded (assuming you can start the conveyor
at least once at full load using the present setup)?
<als>

 

[jraef]

Given that starter configuration, and because the protection circuit you are concerned about is for the feeder to all 3 of the motors, I would agree on the 500msec then. That would avoid the magnetizing current inrush of the motors without creating undue load sharing burden on the first one.

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[Marke]

Hello Ralph

Firstly, I assume that the motors are distributed along the conveyor and are effectively mechanically coupled by the conveyor so that all are spining at the same speed.

If your concern is the start current on your supply by the operation of all starters at the same time, I do not see that this should be a major issue appart from the very fast inrush current to establish the stator field. This is only a couple of cycles and so if you are looking to stagger the starts to reduce this, then 100ms would be fine.

If your concern is not the inrush current, but the start current, then staggering the starters could enhance the problem rather than improving it.

Starting an unloaded conveyor is pretty easy and commonly requires somewhere in the order of 40% torque. You could start the unloaded conveyor on one motor only.
Starting a loaded materials conveyor however, is a different story with starting torques of 160 - 180% being commonly required. In order to start the loaded conveyor, you would need to have all starters/motors working together in order to produce enogh breakaway torque for the conveyor.

If you wish to minimise the start current under a high torque start with a wound rotor motor, it is improtant to keep the slip less than the lip at which the maximum torque is produced, other wise you are beginning to operate the motor in a "high slip" mode and the conversion of amps into newton meters is far less efficient.

Knowing that your application can be required to start a loaded materials conveyor, you must ensure that the initial rotor resistance is high enough to position the maximum torque at greater than 100% slip (I would suggest 120% slip or higher). This will limit the initial start current and ensure efficient conversion of current into torque.
As you reduce the resistance (fluid heating up) the slip for maximum torque will reduce and provided that the rate of change is not too great, and that the maximum torque is sufficient to get the conveyor running, the conveyor will accelerate ahead of the maximum torque movment and reach full speed without causing problems.

I would expect that the start current should remain below 250%. If it increases beyond that, I would suggest that you have a problem with the rotor resistance control.

Best regards,

Mark Empson

http://www.lmphotonics.com

 

[davidbeach]

I don't know what your upstream protection is, but if you have numeric relays with multiple setting groups, you could use a higher setting for the loaded start and then pull the protection settings back down to their normal settings. Allow the temporary overload and then once the conveyor is running again return the settings to the normal settings.

 

[RalphChristie]

Thanks for the comments.

fsmyth: On this moment the start-delays between each starter are between 1 and 2 seconds. In my opinion it is way too long.

Mark: The motors are effective mechanically coupled and are spinning the same speed, although the firstly started motor in effect drag the other two motors during starting (as mentioned by jraef) The longer the delay between the motors, the more the first motor has to work. From an old manual, motors with vapourmatic starters driving conveyors draw approx. 300/350% FLC for max. motor torque. My concern on this moment is to ensure that the initial start current ain't too high (like in starting all three motors at the same time) and also to ensure that the total start-up time is as fast as possible without starting the motors on the same time or overloading one motor.

David: Unfortunately the upstream relays can't be set to go higher during start-up and then to go back to the normal setting. If it was possible however, my concern with a higher setting (although just for starting) will still be that the relay will take longer to trip, especially if there is a fault during start-up.

Thanks again
Regards
Ralph

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[Marke]

Hi Ralph

approx. 300/350% FLC for max. motor torque

If you ensure that the initial rotor resistance is such that the maximum torque is well in excess of 100%, then you are positioning the operation on the "low slip" side of the curve. i.e. to the right of the maximum torque slip. The motor will then efficiently develop torque and accelerate. Provided that the change in resistance is not too fast, the motor will remain on the efficient side of the slip / torque curve and will draw less current.

If the starter is set up such that the maximum torque is at say 80%, then the motor will draw in the order of 300% - 350% current and develop 200-250% torque irrespective of the load. Now position the torque maximum to say 150% slip and the motor will draw around 170% current and produce around 150% torque. (NB - these are not exact figures, just indicative only, but show the trends.)
If the conveyor is empty, it will rapidly accelerate and as it does so, the current will drop and the torque will drop. Acceleration will continue until the motor reaches full speed, or until the conveyor torque equals the torque developed by the motor. The rotor resistance reduces, increaseing the torque at any speed provided that the motor is operating on the right hand side of the torque curve.

If I was designing such a system, I would set up the initial resistance such that the slip is sufficiently high that the initial torque developed by the motor was in the order of 40% and the change in resistance to zero ohms took in the order of 15 seconds.
This would minimise the start current to that required to start the conveyor, it would provide a low torque start for the empty conveyor and a high torque start for the loaded conveyor. I would not expect the start current to ever exceed 200%, and the start that you would get would be very similar to a current ramp soft start mechanically.
The only reason to introduce a delay between the motors is to prevent the inrush current from occuring at the same instant, otherwise the delay should be minimum.

Best regards,

Mark Empson

http://www.lmphotonics.com

 

[RalphChristie]

Thanks Mark.

That makes sense.

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[RalphChristie]

Just spoke today with a mechanical engineer.

His reply was to start the secondary drive first (drive used to take the slack out of the belt) and then the two primary drives. Thus, there have to be a time-delay between motor 1 and motor 2, (the time it takes to tension the belt) but between motor 2 and motor 3 it does not matter mechanically.

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