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Starting A 30Kw Motor By Soft Starter


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#1 AB2005

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Posted 21 January 2018 - 03:30 AM

Hi;

 

Hopping that every one is fine with good health. Many years passed to Log-in. Now need help on an application where we have installed soft starter.

 

We have a trim waste blower at Corrugator machine which has a 380V, 30KW, 4pole AC motor which drives an impeller by the mean of V-belts with double of motor speed. Previously motor was running by star delta starter. As we can not go beyond 6-7 start/stop per hour for such kind of load, so during normal operation of corrugator, if machine was stopped for some reason, trim waste blower was not stopped.

As we calculated that if we stop the blower on each machine stop and start on machine start, we can save energy. So we have installed soft starter 37KW Schneider ATS01N272Q. But we have observed that when blower is start, current goes up to 272A and remains the same for almost 3-4sec, after that current starts dropping and stays at 42A (motor coils connected in delta). The total time which blower takes from zero to max speed is around 6-8sec. We have set the initial voltage around 260v. We always used VFD to control the speed and never used soft starter, so dont have idea about correct operation of soft starter. So, please guide me on following points;

  1. Is above mentioned condition normal?
  2. How many times can we start/stop the motor in an hour by soft starter?

"Don't assume any thing, always check/ask and clear yourself".

#2 marke

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Posted 21 January 2018 - 04:21 AM

Hello AB2005

 

When you use a reduced voltage starter,

  1. you are reducing the torque applied to the motor by the square of the voltage reduction. Half the voltage and the torque is a quarter.
  2. you are supplying instantaneous torque to drive the load at the instantaneous speed. Any surplus torque is converted into kinetic energy, accelerating the motor and driven load to full speed.
  3. If the start torque is too low, the motor can not reach full speed, so there is a torque and current at any speed.
  4. At full speed, the motor and driven load have a kinetic energy that is a function of the total inertia and the rotational speed. During start, you must transfer that energy into the load. If the accelerating torque is low, the load will take longer to accelerate to full speed. If the inertia is low, the load will take longer to accelerate to full speed.

You can shorten the start time, by increasing the start current.

 

When an induction motor is operating, the rotor spins at slightly less than synchronous speed. The difference between synchronous speed and the rotor speed is called the slip speed. The magnetic power from the supply is equal to the torque times the synchronous speed (KW = Speed x Torque). The torque out the motor shaft is equal to the torque times the shaft speed (rotor speed). The difference between the synchronous power and the shaft power is the slip power and is equal to the shaft torque times the slip speed.

During start, the slip changes from 100% (zero shaft speed) to a low value, perhaps 1 - 3% being the full load slip speed.

At 100% slip, there is 100% slip loss which is dissipated on the rotor and as the motor accelerates, the slip losses reduce.

The power dissipated in the rotor during start is considerable, and is at least equal to or greater than the kinetic energy of the driven load at full speed.

 

The maximum number of starts per hour of the motor, is determined by how much energy it can absorb during start, and so is a function of the load inertia and speed.

 

When a reduced voltage starter is used, the slip energy is much the same provided that the motor is able to freely accelerate to full speed. The start frequency with a soft starter will not be higher than with a star delta starter provided that the star delta starter is able to easily accelerate the load to full speed.

 

In addition to the starts per hour limitation of the motor, there is also a start time and frequency limitation for the soft starter.

 

If you wish to increase the number of starts per hour of the motor, then you need to reduce the slip losses. This can not be done by changing types of starter, but a VFD does not allow the motor to operate under high slip conditions and so the rotor dissipation is significantly reduced. The starts per hour of the motor are significantly increased by the use of a VFD as compared to any reduced voltage starter.

 

Best regards,

Mark.



#3 AB2005

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Posted 21 January 2018 - 05:01 AM

Thanks Mark.

 

Its mean, with present conditions, we will have to install VFD to increase number of starts per hour as we can not reduce slip losses.


"Don't assume any thing, always check/ask and clear yourself".

#4 marke

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Posted 21 January 2018 - 05:18 AM

Thanks Mark.
 
Its mean, with present conditions, we will have to install VFD to increase number of starts per hour as we can not reduce slip losses.

Correct

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#5 marke

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Posted 21 January 2018 - 06:28 AM

Check what the motor is capable of. If the start times are very short, then more starts are possible. It is a thermal limitation and the generic figures quoted do not represent you actual conditions.

ten 30 second starts are the same as 20 15 second starts if the start current is the same.



#6 AB2005

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Posted 22 January 2018 - 07:27 AM

On start, initial voltage starts from 235V and current goes up to 255A and returns back to 42A within 10Sec. When motor reaches at almost 80% speed, bypass contactor is on.

 

If we increase the initial voltage 250V, the current goes up to 292A.


"Don't assume any thing, always check/ask and clear yourself".

#7 marke

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Posted 22 January 2018 - 07:34 AM

If you provide this information to the motor manufacturer, they should be able to tell you the number of starts that is OK.






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