SPEED CONTROL OF WOUND ROTOR MOTORS 1

[jmorrish]

We have a GEC / TECO (made by TECO) WR motor...160kW, 6 pole, 450M frame size.

Duty is shiploader luffing winch.

The motor is about 20 yo.

We have had some incident regarding the luff controls and am looking to change the way that it operators to reduce the risks.

The WR motor rotor can be connected to 5 external stage resistor banks.

To go UP ie against the dead weight of the machine - one resistor is shorted and the motor started with 4 resisitors in line, these are dropped out as the speed gathers to full speed.

For the LUFF DOWN motion, we have 2 options.

SLOW SPEED down - this is done by PLUGGING the motor ie all 5 resistors are in series and the motor started in the UP direction ie the motor is trying to go UP but runs BACKWARDS.

FAST SPEED - 2 resistors are shorted and the motor started in the DOWN direction and 2 more resisotrs are shorted for the FULL FAST speed down ie there is ONE resistor still in series with the rotor winding.

A SLOW down speed is required for the luff down movement to provide good control when loading a ship.

The probelm is that the control system calls up a UP control for the luff down motion...this can be an issue when the full upper luff up limit sw is reached and activated.

We would like to remove the PLUG up setup for the SLOW speed down and have a normal run DOWN but still slow speed.

I am trying to come to grips with the wound rotor motor and the speed/torque characteristics.

I rang a man in TECO who told me that the motor will still go to full speed when resistors are in...this is not true. He could not help me further, hence, this email.

From the stuff than I can get my hands on:

A WR motor can/should develop a high starting torque and low stator currents with the resistors in series.
A WR motor is not good at regenerative braking cf squirrel cage induction motors.ie overhauling scenarios.

I have seen sev. articles on the torque from a WR motor at high rotor resistances...ll seem to vary.

What confuses me is:

With all 5 stages in series..the motor can NOT generate enough torque to lift the shiploader > it runs backwards. This goes against item 1 as above....I have yet to measure the resistance of each stage. The only reason for this could be that the resistance is such that the current is that low that the torque (locked rotor) for the WR motor ius not enough to overcome the load. It should be pointed out that the load will vary depending upon the shiploader shuttle extension...further out, more laod on the motor.

For the luff down, I am concefned with over running of the motor and runaway situation...this my be why the last resistor is not shorted for the FAST speed luff down. Can U pls comment on the ability of WR motors to handle overhauling loads.

Further, how can I have a SLOW speed down with the motor actually running (connnected) to run down rather than trying to go UP???


Many thanks for yr help.



James - A mech Bod

 

[Marke]

Hell jmorrish

With a wound rotor motor, it has a synchronous speed equal to the supply frequency etc irrespective of what resistance you have in the rotor circuit.
The addition of resistance in the rotor circuit alters the slip and speed torque characteristics of the motor.

By increasing the rotor resistance, you move the speed of maximum torque further away from the synchronous speed.

Speed control is effectively obtained by moving the point of maximum torque and shifting the intersect point of the motor speed torque curve and the load speed torque curve. This will only work in circumstances where the load torque is constant against speed, or increases with speed.

If you lower the load, you transfer energy from the load back into the supply. The motor acts as a brake and this occurs when the rotor speed exceeds the synchronous speed of the motor. If you have the motor driving forward, i.e. down, then the load will drop at full speed. If you increase the resistance in the rotor circuit, you increase the slip for maximum torque and the motor, which is being over driven by the load, will speed up.
If you wish to use the rotor resistors to control the speed of the motor going down, then you must have the motor trying to drive the load up but with insufficient torque, i.e. high rotor resistance.

If you wish to get a much better result, consider using an active front end Variable Speed controller with the rotor resistors shorted out. This will give you total control over the lifting or dropping loads and return the energy back into the supply.
If you use a standard front end VSD, you will need to used big breaking resistors.

Best regards,

Mark Empson

http://www.lmphotonics.com

 

[jmorrish]

Hi Mark

Thanks for the comments.

As I understand it...more resistance moves the peak/max torque point further away from the syn speed ie to the left and closer to zero rpm.

However, ion our case we start UP with 4 resistors in stage and PLUG down (ie weak UP) with 5 resistors in stage > what gives here...if we have more resistance, ie 5 stages..then surely we have more torque and the PLUG up (ie down with speed control) should not work??? There must be a point where a LOT of resistance kills off the max torque at pull out start??

Can U pls clarify or expand on the drive down case. For full speed down, the last resistor is NOT shorted but the motor speed is about 10-15% higher than full speed up (wih all resistors shorted). I do not understand the reason for not shorting the last resistor if it has no effect on speed control.

I gather from what U are saying that the WR motor will only brake when the speed is higher than the sunc. speed - regardless of the rotor external resistances BUT having more external resistance can be dangerous as the max torque point (to counter the overhauling load) will be further from the sync. point and the motor will overspeed more acc...is this your comment?

Hence, it seems that with a std WR motor, we are stuck with PLUG mode for slow sped control down DOWN motion unless we change to VFD.

If we do change to VFD, are U suggesting that the WR side becomes redundant with rotor circuit shorted out 100% of the time. I have concerns with VFD not been able to generate the required LRT at low freq/starts - they say that the torque is OK but I have seen the use of oversize motors to cater for low speed constant torque applications.

Many thanks


James - a mech bod

 

[Marke]

hello jmorrish

The change in resistance moves the point of maximum torque away from the synchronous speed, but there are two points of maximum torque. There is the maximum torque under motoring conditions (just below synchronous speed) and there is the maximum torque under generating conditions (just above synchronous speed) As you increase the value of the resistance, you shift both of these points symetrically away from the synchronous speed. If the maximum motoring torque occurs at zero speed, then the maximum generating torque will occur at twice synchronous speed.
If you increase the value of resistance above that required for a torque maximum at zero speed, you shift the torque maximum to a negative speed.
The torque maximum is determined by the resistance, reactance and the frequency of the rotor current. If the rotor is being driven backwards at a speed equal to the synchronous speed, then the rotor current will be twice the line frequency. The values of the resistors are selected to place the torque maximum at a useful rotor speed relative to synchronous speed.

Changing the resistance alters the torque curve of the motor. The speed is determined by the motor speed torque curve and the load speed torque curve. For a given load, you can vary the speed by changing the rotor resistance.

For better control, it is possible to use speed feedback to control the rotor resistance and reduct the dependance on the load torque.
You do have a high dissipation in the rotor resistors, the use of a VSD would reduce the energy consmption and give much better control.

Your points:
1. Increasing the resistance moves the point of maximum torque towards the zero speed. Further increase moves the point of maximum torque beyond zero speed and into negative speed. (reverse rotation)

2. Yes too much torque will result in very low torque at zero speed.

3. The motor is operating with reverse rotation. Is speed will be controlled by the position of maximum torque. Increasing the resistance will cause the motor to spin faster in reverse for a given torque load. Reducing the resistance will reduce the motor speed in reverse for a given torque load.

4. Correct the WR motor, (or any induction motor) acts as a brake to overhauling loads, that is loads driving the motor above it's synchronous speed.

5. Agreed.

6. There are some openloop vector drives that can produce over 200% torque without going to a larger motor. Part of the limitation is that most drives are designed with a maximum load capacity of 150%. I know of one drive that can produce up to 400% provided that it is correctly selected and setup. You would not want to run continuously at greater than 100% torque because the motor would not survive, but for acceleration it is OK.

Best regards,

Mark Empson

http://www.lmphotonics.com

 

[jmorrish]

Dear Mark

Thanks again vm for yr comments.

To me, the FAST luff down set up with the resistors in series and been shorted out except for the last one is pointless and should be abandoned - based on yr info, the motor would go to overspeed to suit the torque charactrerics and then slow down a bit as the torque peak moves towards the sync, speed as the resistors are shorted.

In operation, U can hear a definite overspeed and subsequent slowing down - which I thought was the motor catching the dropping load but now may not be that at all.

On the VSD option, the braking torque has to be in excess of the load torque (constant torque application) right down to zero speed.

The s/loader is at the end of a (very) long run from the power to site. The WR motor has the benefits of low inrush currents....hows does the VSD stack up acc. The 160kW motor is the biggest drive other than the 500kW WR conveyor motor for the wharf/yard conveyor.


Many thanks again.


James

 

[Marke]

Hello jmorrish

With a VSD, the motor is never "started", that is it always operates a low slip and therefore does not draw the high starting current associated with motor operated from a fixed frequency supply. The current will change with the shaft torque but will not generally exceed 150% rated current except where it is setup and selected for a very high torque.

If you require accurate speed control down to zero speed, then the best option to use is a closed loop vector system. This will require a tacho or encoder on the motor shaft. Open loop vector drives do not require the feedback but are not quite so accurate around zero speed.

The inverters are also able to contol the brake so that it will not be released until the motor is producing enough torque to hold the load.

I am sure that there will be some experts in your locality who can give you good advise on your installation. If not, there are a number of knowledgable people on this forum to help.

Best regards,

Mark Empson

http://www.lmphotonics.com

 

[Skogsgurra]

James and Mark,

I did some work on overhead cranes a few years ago. There were 200, 110 and 20 + 20 tonnes and travel for four sets of trolleys and two bridges - 10 drives in all. We used the Siemens stator voltage control (thyristors) and kept the rotor resistors.

Have a look at

http://www2.automation.siemens.com/...0a8-9ada-947e8e5dc02f/view/simotras_hd_en.pdf

There are diagrams showing torque/speed relationship with different rotor steps connected.

It is an economical way to get much smother control and the device does the rotor resistance switching for you. Since the motor still is connected to the grid, there are no regen issues. It works the way it always did. Only better.

 

[Marke]

Hi skogsgurra

This is another way of achieving the same thing where the motor is still run in reverse for the dropping load. The thyristor control gives a continuoulsy variable control rather than the steps that are achieved by switching the rotor resistors, and the system uses a closed loop to give the overal performance improvement. I alluded to this earlier but without the stator control. Certainly, this is an option.
I wonder what the cost differential is for this unit as compared to a closed loop vector drive. The closed loop vector drive with an active front end would offer a much improved efficiency, but at what cost.

Definitely worth considering this option for a performance improvement if the cost is OK.

Best regards,

Mark Empson

http://www.lmphotonics.com

 

[Skogsgurra]

I remember that the cost was surprisingly low. I think we had close to a 1 to 2 ratio when compared to a vector control VFD (also Siemens). And you can use a very simple and robust tacho generator instead of an encoder.

 

[cbarn24050]

Hi, a company I was involved with made controllers for this aplication. Thyristor control on the stator, existing rotor resistors kept but now controlled by the drive and a tacho to close the speed loop.

 

[jmorrish]

Hi Mark

re yr earlier reply

The torque maximum is determined by the resistance, reactance and the frequency of the rotor current. If the rotor is being driven backwards at a speed equal to the synchronous speed, then the rotor current will be twice the line frequency. The values of the resistors are selected to place the torque maximum at a useful rotor speed relative to synchronous speed.


What exactly is the forumala for the rotor torque as a fn of the resistance etc??

The reason that I ask is that the load torque can vary depending upon the shuttle postion and the chute loads....I would like to back check that for luff up - NORMAL - that the starting stage is OK for pull out torque...ie would not want to be close to the wind where the first stage of NORMAL UP leads to reverse roation until the next resistor is shorted out. The load torque has increased over the past 20 years due to various mods and load cases.

Many thanks


James

 

[castera]

Hello James

Can you tell me what what kind of Opotor system is fitted, Is it controlled via voltage relays connected across two of the rotor phases or some kind of frequency relays across two of the rotor phases.
This type of system is generally designed so that at full load the hoist will hoist in first notch, with all the four opotor contactors closed ( Near to the star point of the resistance ), these will all pick up at full open circuit rotor voltage and start to drop out at selected voltages as the hoist starts to accelerate (rotor volts falling).
These systems work very well if maintained correctly and are still used on many cranes, you probably need the voltage relays calibrating.
The problem of lowering away from the hoist limit can be overcome by modification to the control circuit, and if you are lucky and have a DC control cct, the insertion of a diode which allows the hoist contactor to pick up when lower is selected should work.

 

[castera]

The resistor is left in circuit so that the torque speed curve for induction generating is flatter and broader, so if the load has a tendency to want run away it will stay within the peak torque of the motor. If the resistor is shorted out, it is (remotely) possible for the motor to accelerate to a point past the peak torque (Above syncronous) and may result in the motor accelerating wilst the torque is reducing.
Try to imagine it in the way that, if you try to start a motor with all the resistance shorted out, you would expect that the motor would start - but with lots of currant, whereas in fact you can get a situation where the motor stalls (with lots of current).
Its sort of the same thing - Its there for a reason, leave it in.

 

[jmorrish]

Thanks Castera

The shiploader is 22 year old....the resistor banks are shorted by time delays in the (upgraded 2000) modern PLC.

I do not know what Opotor is/are.

If U have any suggestions on the resistor shorting - I am all ears.

The shiploader has a shuttle section > hence, the winch (and motor ) load will vary depending upon the shuttle location. I would say that the reason for the last resistor been left in for FAST DOWN is as U say...flatter torque curve.

The main driver for these inquiries was the main upper limit sw. been activated and STUCK on..due to the hard wire interlocks etc..the UP contactor (for PLUG) mode was NOT allowed to come in until the limit switch had CLEARED but the brakes were allowed to LIFT off...the boom started to free fall down as the limit sw. was stuck on.

The boom free fall was controlled by the bck up overspeed devices (and brakes) on the winch drum...last means of defense!!

For this reason, I wanted to see what we could do to prevent this situation from re-occuring.

Before making any changes, I wanted to fully understand the system and configuration/reasoning.

With the real issue in mind, we could re program the PLC /hard wires on the actions resulting from an upper limit sw. activations (incl. sw. monitoring after a drive command) OR when we reach the upper limit....drive down in FAST LOWER straight off the mark and do not go thru PLUG LOWER mode...??? What re yr thoughts?


Thanks


James

 

[castera]

It is common (Not alway's the case) for the service brake to be fed from the motor stator feed. I.E. the dead side of the motor directional contactors - In this configuration the brake cannot be lifted unless you have power going to the motors.
This could be backed up by some form of rotor proving circuit, for instance a voltage relay, fed from rotor circuit. This voltage relay could have a normally open contact in series with the brake relay coil.
In this configuration the brake could not pick up if the motor contactors are not energised. The brake relay will not pick up unless rotor volts are proved, this contact would need to be shorted out once the brake is energised as the relay would eventally drop out as the rotor volts rise.
It is not uncommon to find that some form of protection has been fitted as standard - only to be shorted out due to failure by some well meaning electrician who does not fully understand all the possible consequences of his actions.

Opotor is short for opposed torque.

Lowering away from hoist limit switch.
It is possible to leave the circuit so that fast speed lower is the only way to lower , but this is not desirable in all circumstances, and you need to satisfy yourself that the hoist contactors are allowed to pick up for braking (I am assuming plug reverse braking) if the limit fails to reset.

It should be possible to allow the hoist contactor to pick up via the drivers controller being selected to the lower direction, but again you need to satisfy yourself that the hoist limit switch operation is not compromised.

To try to understand and explain the opposed torque i need to know:
1 What is connected to the rotor circuit ( other than the main rotor resistor.)Is there a bridge rectifier feeding some relays - If so what do the relay contacts operate.
2 How many notces hoist and lower and for how many notches lower are the lower contactors energised (I am guessing only the last notch).
3 When you say five resistors - do you mean five resistors per phase.
4 Is the star point shorted or are there contactors which pick up to short the star point.
5 Are the contactors at the star point end of the resistor singular or mechanically tied pairs.