Jump to content


Photo

Locked Rotor Current


  • Please log in to reply
12 replies to this topic

#1 bob

bob

    Member

  • Full Member
  • PipPipPipPip
  • 187 posts
  • Gender:Male

Posted 13 September 2005 - 04:15 AM

Hi,

How important is the LRC when selecting a motor for particular application ?

Bob

#2 marke

marke

    Posting Freak

  • Moderator
  • PipPipPipPipPipPip
  • 2,642 posts
  • Gender:Male
  • Location:Christchurch, New Zealand

Posted 13 September 2005 - 06:25 AM

Hi Bob.
It depends on your application and your requirements.

If you are using with a VFD, then I would ignore the LRC.
If you are using a full voltage starter, then the LRC is important to the extent that it is the start current that will be drawn from the supply.

If you are using a reduced voltage starter, the LRC will help to determine the lowest start current that can be achieved.

Have a look at the motor control and motor starting pages on my site at http://www.LMPhotonics.com

Best regards,

#3 bob

bob

    Member

  • Full Member
  • PipPipPipPip
  • 187 posts
  • Gender:Male

Posted 13 September 2005 - 06:43 AM

Hi Marke,

Thanks for your prompt and always intelligent reply. I got the point. What is the relation between LRC and the starting current.I agree that in the case of a full voltage start the LRC will determine the maximum current that will be drawn from the supply. But most manufactures quoted LRC at a value of five times FLC. This means that at full voltage start, the starting current will exceed the LRC.

Bob

#4 GGOSS

GGOSS

    Senior Member

  • Moderator
  • PipPipPipPipPip
  • 407 posts
  • Location:Melbourne

Posted 13 September 2005 - 06:49 AM

Hello Bob,

It really depends on the application and more importantly on capacity of the power supply .

A motor will draw Locked Rotor Current at the momment of 'power on' when started via a Full Voltage/Accross the Line/DOL starter. Under full voltage starting conditions the motor will draw LRC irrespective of whether it is started bare shaft or fuly loaded. The addition of load to the motor shaft simply increases starting time.

On installations where the power supply is weak/limited, the use of a motor with a high Locked Rotor Current may cause an unacceptable voltage drop to occur. This may adversely impact other equipment connected to the same network.

One simple calculation I like to do when comparing motors of different manufacture in order to assess application compatability is to divide the value given for Locked Rotor Torque by the value given for Locked Rotor Current for each motor. The motor which produces a result closest to 1 will generally provide the best starting performance, particularly under reduced voltage starting conditions. I then proceed to calculate how much torque will be delivered at the maximum allowable level of starting current and if that proves to be greater than the torque required by the load, I know I can proceed with that particular motor. NB; By best starting performance, I am refering to the level of torque produced by the motor for a given level of current draw from the supply.

I am not certain that the above fully answers your question. If not feel free to ask for more information, and we'll do the best we can for you.

Regards,
GGOSS

#5 bob

bob

    Member

  • Full Member
  • PipPipPipPip
  • 187 posts
  • Gender:Male

Posted 13 September 2005 - 07:21 AM

Hi GGoss,

Thanks for your reply. My point particularly concerns weak power supply line. In the case of full voltage start, the starting current is around seven times the FLC and as starting current>LRC, the starting current should be the determinating factor for the calculation of voltage drop.
Is there any relation between the starting current and the LRC ?

bOB

#6 marke

marke

    Posting Freak

  • Moderator
  • PipPipPipPipPipPip
  • 2,642 posts
  • Gender:Male
  • Location:Christchurch, New Zealand

Posted 13 September 2005 - 09:45 AM

Hello Bob

When you use reduced voltage starting, the minimum start current is dependant on a number of factors.

  • Speed / Current curve of the motor
  • Speed / torque curve of the motor
  • Speed / torque curve of the driven load
  • Starter performance.

      The driven load determines the minimum starting torque. The motor must develop at least that torque in order to start the driven load. If the torque is less than ther required torque, the load will not reach full speed.
      The motor converts amps into newton meters (start torque). The starting efficiency of motors varies dramatically. Ideally, the motor should have a low locked rotor current and a high locked rotor torque. When you reduce the start voltage applied to the motor, you will reduce the start torque by the voltage reduction squared.

      If you download the Electrical Calculations software from http://www.lmphotoni...om/busbar32.zip you can see the effects of different motor curves an a given load etc.

      The full voltage start current always equals the LRC when the rotor is stationary. The start current remains high untoil the motor approaches full speed.

      The LRC of modern motors is commonly 700 - 850% of rated current.

      Best regards,

#7 jraef

jraef

    Posting Freak

  • Moderator
  • PipPipPipPipPipPip
  • 683 posts
  • Gender:Male
  • Location:USA, California

Posted 13 September 2005 - 03:22 PM

My opinion is that there is a semantics and measurement issue going on here. With an Across-the-Line (DOL) start, starting current and LRC are essentially the same, but the semantics makes all the difference, and how you measure it may skew your results.

Locked Rotor Current is a specific engineering definition meaning the peak current drawn by the supply at rated voltage when the rotor is not moving. That could take place at startup or while running (jam condition), the definition makes no distinction.

Another term not mentioned here but often misconstrued is Inrush Current, which is the current drawn by the stator windings instantaneously when power is applied in order to establish a magnetic field in the core. This can be much higher than LRC (ie.e 6 - 20 x FLC) because the flow of current is restricted solely by the winding resistance, prior to any induction taking place. It is, however, very short in duration, typically 1-1/2 to 3 cycles.

Starting Current is a determined value specific to the application, and takes many more factors into the equation, such as Inrush Current, LRC, starting method, supply capacity, load etc.

So depending on your measurement method, the Starting Current may in fact appear to be higher than the LRC simply because your measuring device is capturing the Inrush Current. For instance, your motor may have a nameplate LRC of 600% FLC, but the Inrush Current may be as much as 1000% FLC. A metering device with a fast enough sampling rate may integrate that 1000% into the peak calculated value, throwing it off and making it appear higher. In reality, if you have a weak supply, that can can cause the the line voltage to drop, and that in turn will then decrease the current draw just the same as a reduced voltage start would, so the average starting current will actually be lower.

So in answer to your question, yes there is a "relationship" between Starting Current and Locked Rotor Current, because LRC is a component of Starting Current, but that is only part of the equation. You cannot apply a specific formula that could help you to pre-determine anything useful with only that value.
"He's not dead, he's just pinin' for the fjords!"

#8 GGOSS

GGOSS

    Senior Member

  • Moderator
  • PipPipPipPipPip
  • 407 posts
  • Location:Melbourne

Posted 14 September 2005 - 02:28 AM

I've heard people talk about differences in starting current and locked rotor current under full voltage starting conditions but have always worked with LRC only when performing volt drop calculations.

To my way of thinking the motor at rest (at momment of power on or when jammed) presents as a fixed impedance and that determines how much current will be drawn from the supply.

I may be wrong but am am yet to be convinced there is a difference between starting current and locked rotor current and suspect that those who are measuring a difference are simply experiencing overshoot on their metering equipment.

As I say, I may be wrong (it does happen occassionally) and therefore I would welcome any further comment on the subject.

Regards,
GGOSS

#9 bob

bob

    Member

  • Full Member
  • PipPipPipPip
  • 187 posts
  • Gender:Male

Posted 14 September 2005 - 03:56 AM

Hi all,

Am I right to say that irrespective of the starting mode, the LRC should be considered when calculating volt drop on power system.

Bob

#10 jraef

jraef

    Posting Freak

  • Moderator
  • PipPipPipPipPipPip
  • 683 posts
  • Gender:Male
  • Location:USA, California

Posted 14 September 2005 - 04:55 AM

No, not at all, the starting method can make all the difference in the world when it comes to voltage drop. That is one of the reasons why soft starters exist! Voltage drop is caused by current draw and system power availability. Since you usually cannot control the system power supply, the only solution is to reduce the starting current. Reduced voltage starting specifically does that. I think I left that unclear in my earlier post, sorry. Starting current and LRC would be the same for DOL starting, but not for any other method of reduced voltage starting (well, maybe Start-Delta is problematic, but that's another subject).

When you reduce the voltage, you reduce the current drawn by the motor. If you reduce the current, it will help to reduce the effect that the starting current has on your power supply.
"He's not dead, he's just pinin' for the fjords!"

#11 bob

bob

    Member

  • Full Member
  • PipPipPipPip
  • 187 posts
  • Gender:Male

Posted 14 September 2005 - 05:29 AM

Thanks jraef. But that bring me back to square one that is LRC is immaterial when selecting motor for reduced voltage , soft start and VSD applications.

Bob

#12 marke

marke

    Posting Freak

  • Moderator
  • PipPipPipPipPipPip
  • 2,642 posts
  • Gender:Male
  • Location:Christchurch, New Zealand

Posted 14 September 2005 - 06:01 AM

Hello Bob

When I am involved in recommending or selecting motors for reduced voltage starting (usually soft starting), I first ask for the speed torque curves for the load, and the speed torque and speed current curves of the motor. This enables me to determine the required stating current for the motor and machine and from this I can calculate the voltage drop.
If I can not get the curves, I then use the LRC and the LRT to determine the best motor, and use these figures to estimate the start current requirements.
If you are looking at a number of motors and wish to select the best motor, I take the LRT % and divide this bye the LRC % and go for the motor that has the highest number. This will be the motor that will give you the lowest start current.

So, in answer to your question, I use the LRC and LRT figures of the motor to select the best motor, then use these figures to determine that starting current. The start current is then used to calculate the voltage drop.

Best regards,

#13 GGOSS

GGOSS

    Senior Member

  • Moderator
  • PipPipPipPipPip
  • 407 posts
  • Location:Melbourne

Posted 16 September 2005 - 01:09 AM

Hello Bob,

to further illustrate what we are talking about in the above, take for example the following two motors:

Motor A: LRC = 650%FLC, LRT = 180%FLT

Motor B: LRC = 740%FLC, LRT = 120%FLT

Assuming we want to start a centifugal pump requiring a starting torque of 35% via a soft starter, we calculate starting current as follows;

For Motor A: Ist = 650% x square root 35%/180% = 650% x 0.44 = 286.6%FLC

For Motor B: Ist = 740% x square root 35%/120% = 740% x 0.54 = 399.6%FLC

Therefore both LRC & LRT are very important considerations particularly when attempting to start a load on a weak/limited supply.

As you are already aware, the level of voltage drop that occurs on a supply network during motor starting is directly related to the starting current drawn from the supply. From the above it should be apparent that Motor B will cause significantly more voltage drop than Motor A.

Hope that helps to clear things up.

Regards,
GGOSS




0 user(s) are reading this topic

0 members, 0 guests, 0 anonymous users