Voith do not know how electrical and mechanical combinations work.

First, you need to control the inrush current to below 600%. If you use a primary voltage regulator for the motor MCC, you can control the rate of rise of RPM to the level where the inrush drops off dramatically. Usually about 80-90 % of full speed.

So the trick is to accelerate the motor to near full speed over ~3 seconds, while not accelerating the conveyor. The fluid coupling will just begin to move the conveyor. The major mass motion should just begin to take root to tax the coupling and motor in about 4-5 seconds. In the mean time the motor has passed the difficult inrush stage, while the fluid coupling torodial shear work is just getting started to move the conveyor.

Take care and do not overfill the coupling. High fill levels can kill the startup when the fluid coupling pump curve crosses over the motor torque vs RPM signature curve.

If you need more help, a bottle of excellent red is necessary. ■

We can model these details explicitly. You give the Voith model, WEG torq vs RPM and in-rush current curve. This will be more than one bottle's worth.

It will take maybe two days to complete, given the conveyor is not complex, and you have all general arrangement details and equipment component specs, and the design is not bogus. ■

Here's a couple of extra thoughts -

1) look for a motor from another supplier which may have a lower locked rotor starting current.

2) consider a traction fluid coupling with a delay chamber and use a star-delta start. I did have some calculations, but they were done in the mists of time and I've no idea what happened to them so it's just an idea for you ■

Dear Alex,

You have a designed and calculated drive.

A starting time calculation must have been made, using the torque curve of the motor, the fluid coupling curves and the involved inertia and loads.

The electric switches, cabling, overload relays and their settings are part of this calculation.

As far as I know, the inrush current is the current peak to build the magnetic field and can be as high as 20 times, but only for a few milliseconds.

Depending on the starting time, the cable will be overloaded for this starting time.

The cable can withstand 600% nominal current for a certain time and may be 760% for a shorter time.

As long as the starting time is shorter, this should be no problem

Applying a soft starter reduces the starting current and the starting torque, thereby lengthening the starting time.

Using the soft starter, a new toque curve for the electric motor is generated and using this torque curve as input in the above-mentioned calculation gives you the new starting time.

This new starting time and the slip of the fluid coupling during that time, determines the heating up of the coupling and the supplier must be able to confirm the suitability of the coupling. (Or not)

A Star-Delta start cannot start under full load.

I have a simple spreadsheet for an acceleration time calculation for DOL Star/Delta and soft starter.

If you are interested, I can attached the file to a post reply.

Have a nice day

Teus ■

Dear All,

Aceleration time of the conveyor is not the problem. It will be sufficiently quick to keep the thermal fuse from discharging.

The problem is to acelerate the motor without encroaching on the high inrush current. The primary voltage regulator can keep the inrush from exceeding 600% by gating the voltage to a lower than nameplate value while getting the motor to near full speed where the motor demand current will fall dramatically. As I said the motor will need to be near or exceed 80% of full speed. The fluid coupling will allow the motor to accelerate and not the conveyor.

The allowance needs to be studied the set the voltage vs time ramp, for this we need to see the conveyor, coupling and WEG motor specs. IF the voltage is gated too quick, then the inrush will exceed the spec. IF the voltage is gated too slow then you heat motor and fluid coupling beyond their acceptable limits. ■

The motor is to drive a "reclaimer feeder". What type of machine is this, a high inertia load or a low inertia load?? ■

Dear Nordell,

Locked rotor current and inrush current are two different phenomena.

In addition, I believe that an electric cable, not capable of starting an electric motor DOL, is chosen according to the governing rules.

Run an extra cable along the existing one, perhaps?

BR

Teus ■

Originally Posted by Teus Tuinenburg

A Star-Delta start cannot start under full load.

I have a simple spreadsheet for an acceleration time calculation for DOL Star/Delta and soft starter.

If you are interested, I can attached the file to a post reply.

Have a nice day

Teus

The spreadsheet would be interesting to see ..... ■

Gear designer,

Here it is.

The yellow cells are for input.

success

Teus

Attachments

accelerationtimecalculation (ZIP)

■

Thanks Teus I need to study it a bit more.

My thinking was along these lines ---

1) lets assume the feeder has a high starting torque requirement

so the machine won't start running until a significant torque is being generated at the output of the fluid coupling.

2) lets have a fluid coupling with a significant delay chamber.

so when you apply power to the motor the acceleration is initially controlled by the combined inertia's of the motor rotor and the input half of the fluid coupling.

3) lets start in star to limit the locked rotor current and ignore the fluid in the fluid coupling.

so the motor will start and accelerate to full speed, the time being controlled by the combined inertia seen by the motor.

4) but it's not actually like that, as the motor accelerates fluid bleeds into the hydraulic circuit generating torque and reducing the acceleration of the motor. Eventually the hydraulically generated torque will limit the speed of the motor as being in star limits the torque the motor can generate.

5) the question now is what speed will the motor have reached in star, and this must be a function of the rate of bleeding the fluid into the couplings hydraulic circuit. The slower this happens the higher the speed that the motor achieves.

6) according to some motor data I have for a 110kW 1480 rev/min motor the current drops to 600% FLC at around 500-700 rev/min. If the characteristics of the fluid coupling would allow this to speed to be achieved in star, then at this point the motor could be kicked into delta to do the final acceleration and start the feeder.

7) so start in star, monitor the motor speed, change to delta at say 600 rev/min.

Of course this assumes that there is a suitable fluid coupling and that would have to be discussed with the coupling manufacturers technical department (certainly NOT their salesmen )

Please feel free to point out the errors in my logic, I won't be offended as I am writing this on a Saturday night after some Pear Cider ■

Originally Posted by alex-kaveh

Dear All,

For one of our reclaimer feeder machines I have selected E-motor with following specifications:

110 kW

Locked rotor current of 760% of full load current

and also I have used a Voith fluid coupling between e-motor and gearbox.

Our client now says that their cabling can support just to max of 600% of full load current in the start-up.

I talked with WEG comany and they suggested using soft-starter for the e-motor and fluid coupling as already selected and mounted on the machine.

On the other hand I talked with Voith company experts and they say because you are using fluid coupling there is no need to use soft-starter. as it will over-heat the fluid coupling in the start-up.

Can you please share your ideas if you have any experince with fluid couplings and saft-starters?

Thank you in advance,

Dear Mr.Alex,

I feel M/s.Voith is correct. We removed fluid couplings wherever VVFD ( Soft drive?) installed. They are running OK. Thanks to Mr.Desinger's advice I modified the speed sensors so that driven pulley will have coordination with drive pulley.

Rgds,

Rgds, ■

Any thing electrical to reduce the current and increase the start up time will increase the heating of the motor and this may effect the number of retarts per hour.

Note that you can not rely 100% on motor spec and data eg 760% may in fact be higher.

The fluid coupling will allow the motor to get to speed very quickly and then the fluid coupling will address the starting of the load.

We had a soft starter on a belt conveyor that was not sized to suit thermal heating - the motor ran out of thermal capacity during a loaded start and shutdown. With this arrangement - the motor has to accelerate the motor plus load together. With the fluid coupling - the intertias are split.

JM ■

Dear designer,

What you are describing is correct.

see:

http://www.voithturbo.com/applicatio...stant-fill.pdf

For a star/delta arrangement, 3 extra cables have to be laid from the schwitchboard to the motor.

These cables can also serve to reduce the Amps in the existing cables, which was the only problem. Installing the switch gear is then not necessary anymore.

Calculating the acceleration time of the whole system becomes with a fluid coupling more complex.

If you assume that the electro motor is only loaded by the fluid coupling (irrespective of the outgoing shaft rpm of the fluid coupling), then the acceleration of the e-motor is easy to calculate.

The acceleration of the installation after the fluid coupling then becomes a second calculation with the transmitted torque of the fluid coupling as the accelerating torque and the installation load and inertia to be overcome.

Actually, 2 calculations, whereby the e-motor is coupled to the fluid coupling and whereby the fluid coupling is coupled to the installation.

When the fluid coupling torque curve is a function of the rpm difference ingoing and outgoing (torque converter) it becomes really complicated.

have a nice day

Teus ■

Dear Alex,

1)Correct. Interaction between the 2 systems must be investigated.

2)The new e-motor curve with a limited starting current at 600% can be determined from the original torque-rpm curve of the e-motor, using my spreadsheet.

3)Correct. However, including the installation load and reduced moments of inertia.

Why make all the cost to reduce the cable current, when a simple extra cable can do the same?

All for now

Teus ■

Dear Teus,

Please inform me how the inrush and locked rotor currents are different, assuming the inrush must start at the same point. I have maybe mistakenly believed inrush current is defined over the full motor speed curve, where locked rotor is just the starting point. ■

Dear Nordell,

Consider a squirrel cage e-motor as a transformer, which is short circuited on the secondary side. (The short circuited squirrel cage).

When the voltage is switched on, then the frequency on that transformer is the grid frequency. (50Hz or 60Hz)

The induced current in the squirrel cage counter acts the current in the stator windings, limiting the primary current at the short circuit current. This is the locked rotor current.

Before any voltage is induced in the squirrel cage, there must be a magnetic field.

Without the existence of a magnetic field the only resistance is the Ohm resistance and the starting current is then calculated as I-inrush = V/R-Ohm.

This inrush current is used to build up the magnetic field in the motor.

This current is depending on the exact moment when the Voltage is switched on in relation to the phase moment. However, in a 3 system, there are always 2 phases carrying more than 0 Volts.

When the magnetic field is generated, which takes only a few milliseconds, the counter acting induction is generated and the current drops to the locked rotor current.

The locked rotor current is calculated as: I-locked rotor = Voltage / Impedance

Impedance = SQRT(omega^2.L^2 + R-Ohm^2)

Omega = 2*pi*Frequency/60

Frequency = 50Hz or 60Hz

The induced rotor voltage is a constant * slip frequency and the resulting current is:

I-rotor = constant* slip frequency/Rotor impedance

When the rotor is increasing in rpm, the slip frequency is decreasing (0 at synchronous rpm) and the rotor current is reduced as well.

Seeing the electric motor as a transformer, implies that the primary current also decreases.

(disregarding energetic losses, 0 at synchronous rpm)

In 1992 I was involved in the design and building of a 400 tons/hr unloader, where the thermal/maximal relays of 132 kW motors (approx 190 Amps) activated on the inrush current of approx 2300 Amps, when started. (But not every start).

The locked rotor current was about 1300 to 1400 Amps.

Start up time was about 30 seconds (screw compressors and calculated with the spreadsheet)

Replacing the safety relays by relays that were able to withstand >2500 Amps inrush current solved the problem.

All for now

Teus ■

Dear Teus,

Thanks for the detailed explanation. We need a little more enlightenment.

When we purchase large motors we received a certificate of the inrush current associated with the motor rotor-stator design from the manufacturer. The inrush currect shows a maximum of ~700 - 800% x full load operation at the lock rotor point. The manufacturer's inrush currect maintains a significantly high value over ~80% of the rpm range before falling off to the nameplate current (power) rating.

I believe you are suggesting this motor signature inrush current differs from the observed value., a la cable size et al.

If we drop the voltage, to say, 20% of grid value, to engage the motor, do we drop the inrush current (motor manufacture or observed)? We do need enough voltage to push the current to the motor to get rotation.

As you note, if we control (drop) the voltage fed to the motor, via the SCR-type starter, the locked rotor current also drops. When we apply the fluid coupling, why does this not achieve the goal of allowing the motor to pass through the high inrush current stage, at a significantly reduced level, achieving >80% motor speed, before taxing the drive to accelerate the conveyor mass? This motor acceleration phase is very short compared to the conveyor acceleration phase.

Eager to learn more. ■

Dear Nordell,

One problem seems now solved.

Where your manufacturer uses the definition “inrush current”, I used the definition “locked rotor current”

May be, you can ask your supplier for the definition for my “inrush current” or magnetizing current.

The torque, delivered by the motor is a function of (motor Voltage)^2

Lowering the motor Voltage by 20% reduces the torque to 0.8*.08=0.64 times the full voltage torque.

In star/delta, the voltage is lowered to 1/(SQR(3)) and therefore the torque decreases to (1/(SQR(3)))^2 = 1/3

Lowering the voltage can be achieved by a soft starter or resistors, as in the past with a wound rotor.

If you keep the Volt to Hertz ratio constant you will have a speed regulation at constant torque.

If a power supply cable is sized according to the standards (wherever on the world), the problem of an undersized cable (as in this thread) is impossible.

All regulations allow for direct on line starting, whereby the cable temperature stays lower than the motor temperature, safeguarded by overload relays.

In case of long starting times and if you are in doubt, every engineer would calculate the involved temperature rise of the cable in a short circuit calculation.

Calculating a drive train is a matter of equilibrium of torques at any moment of starting.

If the e-motor is not capable of supplying enough torque through the fluid coupling, which is in fact reaching the required rpm (torque transmittal is function rpm), the coupling heats up.

A calculation in the time domain during startup accounts also for the dissipated heat in the coupling and will tell, whether the arrangement is OK.

Engineering can be fun to do, when the suppliers provide the necessary and to the point information.

In these cases it is an advantage to be an electrical and a mechanical engineer.

All for now

Teus ■

A message to Alex Kaveh.

If you have followed all that has been said here regarding squirrel cage 3 phase motors you will have learned invaluable lessons regarding this type of motor.

In the whole of bulk materials handling squirrel cage 3 phase motors are the most commonly used motive unit. I regards them as the most simple yet elegant of machines, but frequently engineers are ignorant of their principles of operation.

I always prefer fluid couplings to electronic soft starts when using squirrel cage 3 phase motors, but maybe I'm just old fashioned. What is essential when designing a drive unit is to understand the characteristics of the machine to be driven, typically is it a high or low inertia machine has it a high or low starting torque requirement, to understand the characteristics of the motive unit, and to understand the characteristics of any intermediate units, like a fluid coupling or a soft start device. Understanding will lead to the correct selection of components, and the ability to solve problems when a drive system is not performing as expected. ■

Here's something I've picked up from another forum. I'd never heard of this before. Comments anyone??

When a star delta starter is employed, there can be very high current and torque transients occur during the start.

The problem is two fold.

1) During start, the motor impedance is low, independent of load, until the motor almost reaches full speed. If you change over from star to delta at less than about 85% speed, you will draw almost locked rotor current, resulting in a high start current until the motor reaches full speed.

2) While the motor is connected in star (wye) you have a rotating magnetic field in the stator which is rotating at line frequency. This induces a current in the rotor and develops a torque field at line frequency. When you open the star contactor, you have a magnetised rotor spinning at less than line frequency. The effect of this spinning magnetised rotor inside a stator is that of a generator. The frequency is related to the speed of rotation and is not synchronised to the supply. When you close in to delta, you are closing in on a non synchronised generator so there is a current transient due to the vector addition of the generated voltage and the line voltage. This current establishes a new field in both the stator and the rotor and once the field is established, the current is limited by the effective impedance of the stator at that speed.

The current transient can be greater than 2 times the full voltage start current, and the torque transient can be greater than 4 times the full voltage start torque.

The result is more damage to the supply and the driven load than a Direct On Line (Across the line) or full voltage start. ■

Originally Posted by alex-kaveh

Thank you all for your valuable replies. I really appreciate it.

To sum up and also clarify what we discussed:

1- I understood that we can use both fluid-couplings and soft-starters in one single drive (in general). So the idea that it is not possible to use both in single drive is not correct.

.................

Thank you again all.

When we have VVVF, why do we need fluid coupling? Fluid coupling maynot work below certain RPM of motor.

Rgds, ■

Dear designer,

I suppose that the ignorant engineers, you are referring to, are the “overall” engineers, who are designing complete systems.

Each component in that system requires in depth knowledge and to make a working complete system, it is necessary to understand the working of each component and there interaction in the system, before the whole system works.

To prefer one system over another, depends on the application and the economics.

The energy efficiency of fluid coupling is less than that of an electronic soft starter.

Thinking about the application of the fluid coupling, it becomes more and more complicated.

A weak fluid coupling results in a fast start up of the e-motor and a slow start up of the machinery. The fluid coupling slip is then high.

A hard fluid coupling results in a slower start up of the e-motor and a faster start up of the machinery. The fluid coupling slip is then low.

The stiffness of the fluid coupling matters to what the e-motor sees as load and inertia.

Calculating this “feed back” function of the fluid coupling in the drive train is a real engineers job.

There must be engineers out there (coupling manufacturers), who have done this.

And, how are all the fluid couplings in the world calculated?

The effect of star delta switching that you picked up (it seems that you are interested) is correct and in line with what I wrote earlier in this thread.

The description given, relates to the inrush current, related to magnetizing.

When switched to delta, the peak current lasts shorter than in DOL.

Sometimes, at start up of a complete installation, it appears that the electrician connected the star in the right rotation direction and the delta in the opposite direction.

The result is similar, but much more violent and costs at least 2 fuses.

Dear sganesh,

Your remark is exactly the reason, why engineers must understand their profession and do the calculations before rather than after they install the equipment.

Have a nice day

Teus ■

Originally Posted by sganesh

When we have VVVF, why do we need fluid coupling? Fluid coupling maynot work below certain RPM of motor.

Rgds,

Where did the variable speed come from???

We were talking about soft starts i.e. variable voltage starting. ■

Dear designer,

During the start of an electric motor, there is a variable speed.

A Volt-Frequency regulator can start at maximum and constant torque.

A bit expensive may be, but it works.

All the best

Teus ■

Teus,

In English there is a difference between frequently engineers are ignorant of their principles of operation and ignorant engineers, the former being ignorant in a small area, the latter being brain dead.

My problem with soft starts comes from experience of them being retro-fitted to my machines.

For example some years ago on a bucket elevator that wouldn't start after a fluid coupling was replaced but blew it thermal overloads at each attempt. The solution was a high starting voltage with a very short run-up time, so ending up with a start that wasn't very soft!!

And again recently where we recommended a soft start unit and the client engineers decreed that the current limit be set to a maximum 200% of full load plate amps, then complaining of erratic starting!! Had to argue forcefully that they needed to increase the current limit during the starting phase to ensure there was sufficient starting torque.

The reason I tend to favour fluid couplings is that in my formative years there were no electronic "soft start" units! Hydro-dynamic couplings are relatively easy for a mechanical engineer to understand whereas soft starts and variable frequency units tend to be "boxes of black magic".

No-one can really design a drive without a knowledge of the characteristics of the motive unit, the drive components used and the machine being driven.

Interesting that you comment on going from star forward to delta backwards, we supplied a couple of mobile machines that ran for many years where the change of direction was simply to direct reverse the motor at each end of it's travel and there was never any problem ■

Originally Posted by Teus Tuinenburg

Dear designer,

During the start of an electric motor, there is a variable speed.

A Volt-Frequency regulator can start at maximum and constant torque.

A bit expensive may be, but it works.

All the best

Teus

My concern was that reference to a variable frequency unit to drive the motor was a confusion as all that was being discussed was variable voltage starting.

Variable frequency is another whole can of worms!! ■

Dear designer,

I understand the nuance of the type of engineers.

Experience builds knowledge and in the case where you recommended the soft start, the client engineers should have been convinced by an acceleration calculation, unless they belonged to the latter type.

I do not see the modern electronic equipment as “black boxes” (someone made them).

I see the new electronic equipment as a much more flexible replacement for reliable and proven, but energetic inefficient components.

New technology should be embraced by old and young engineers.

My example was not meant to reverse the motor, but to complete the start through a star/delta diagram that was mistakenly wired in the wrong way. (Two phases crossed in delta compared to the diagram in star).

Now back to the original thread.

Dear Alex,

1)If the cable is the limiting factor, then the soft starter is not the only option. If the switchgear and the safety relays and/or fuses or the powersupply itself are the cause, then it might be the best solution.

2)A fluid coupling also absorbs shocks in the system and as we have established, the fluid coupling characteristics play a role in the start up. A calculation can reveal the consequences. (Is Voith not directly replying?)

3)Starting the e-motor within 3.5 seconds is fast. Use my spreadsheet with the fluid coupling's curve as load and only the inertia of the e-motor and the first half of the coupling. Do the same with a soft starter set at 600% current limit and verify, if the start up time is within 3.5 seconds. Obviously, Voith wants to limit the heat dissipation in the coupling.

4)Why not investigate the necessity and calculate the starting up event before ordering?

Alex, did you receive my e-mail address?

Do you have the nameplate data and the torque and current curves of the e-motor?

In case of a standard motor from a well know manufacturer, the curves are also standard.

Are the fluid coupling curves available?

Have a nice day

Teus ■

Teus,

Can you confirm that there is a square law relationship between the speed of the input half of a fluid coupling and the torque it generates when the output half of the fluid coupling is stationary (100% slip condition)? ■

Dear designer,

I would say, yes.

The oil in the coupling is accelerated and the kinetic energy is then converted into mechanical energy, which is radial speed * torque.

The torque is proportional to 1/2 m v^2 and the energy is proportional to 1/2 m v^2 * v = 1/2mv^3.

Propeller law as for ship propellers, centrifugal pumps and - fans.

BR

Teus ■

Dear Experts,



What is the minimum acceleration time desirable for the conveyors, i.e., ramp up time ( the drive pulley RPM will start from 0 to achieve its fullest designed PRM for the conveyor)?

Normally below 30 KW fluid couplings are not used ( Am I right ? ).

How this ramp up time is designed?

Regards, ■

Dear sganesh,

Selection charts of fluid coupling manufacturers indicate the availability of fluid coupling from 0.2 kW at 500 rpm and up.

A ramp up time is designed like any other design, by calculation.

The design calculations take care of strength limits, torque limits, heat limits, investment limits, safety limits, etc

Regards

Teus ■

The use of a fluid coupling, or not, is down the the machine designer, his opinion and experience.

The ramp up time is a function of the conveyor, it's starting friction and it's inertia, and the torque applied to it from the drive assembly. It is calculated from these inputs. If you don't like the result change the inputs until you get results you like. Of course you may not be able to modify the conveyor so then you're just down to changing the drive. ■