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Inrush current deruction 2

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aprilee

Electrical
Mar 15, 2006
10
I need to install a soft-start on a 200HP, 460v, 226 FLA motor. My question is, what would a 30% voltage reduction do to the inrush current? It would reduce it by how much? Thanks!
 
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The rough approximation says that you get a 30 percent reduction in starting current. You will get a near 50 percent reduction in starting torque. (1.00-0.30)^2

Gunnar Englund
 
Thanks a lot for the answer. That answers my question for now. Do you know the formula for the inrush current? Thanks!
 
In a soft starter, current reduction follows voltage reduction pretty much 1:1 until the motor accelerates to 90% speed or greater, and torque is always varied by the square of the applied voltage.

"Inrush current" is technically the magnetization current when power is first applied to the windings, and can be anywhere from 800 - 2000% FLA for 1 cycle or less. You probably meant "Starting current", which is also defined as Locked Rotor Current (Locked Rotor Amps). LRC/LRA is typically 500-600% of FLA, but is specifically determined by the NEMA Locked Rotor kVA Code letter for any particular motor design.

Eng-Tips: Help for your job, not for your homework Read faq731-376 [pirate]
 
Thanks for the answer. You are right, I mean LR current. So, this 1:1 reduction must affect the LRA right? Do you know how long does a typical motor take to get to it's rated speed? Thanks!
 
100% related to the load on the motor. There are very complex software programs that can answer that question, but you need a lot of information such as motor inertia and torque-speed characteristics as well as load torque-speed requirements and inertia etc. One of our regular contributors to this site, Marke, sells a handy little inexpensive software program that will give you a good idea with minimal information.


Other than that, I use SKM, others here use ETAP, both of which would set you back $5K or more if you wanted definitive proof of success before you buy a starter.

Eng-Tips: Help for your job, not for your homework Read faq731-376 [pirate]
 
It's a centrifugal ash booster pump.
 
Strictly from my personal experience, with no knowledge of your particular pump, head, power source, environmental conditions or anything else, I say at 350% current limit it should start in 10 seconds, more or less.

Then again, it could be 20 seconds or more, could require 400% or 450% CL, or may not start at all if your power source is too weak or the transformer is too small or the head is too high or the distance to the motor is too long or the motor is undersized etc. etc. etc. It's kind of like asking "can I drive from San Francisco to Los Angeles on one tank of gas?" without saying what kind of car, how many passengers, which road, how fast, wind conditions, etc.

Eng-Tips: Help for your job, not for your homework Read faq731-376 [pirate]
 
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