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Short circuit and Arc Flash motor's load factor contribution?

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majesus

Electrical
Aug 16, 2007
262
I am studing short circuit and arc flash analysis and I am using Etap to do some experiments. I noticed that when determining the short circuit current value whether it was 3 phase bolt, L-G, arc flash, etc, a motor's load factor did not contribute to the fault current. In other words, if the motor is loaded to 40% or 90% it does not make any difference for short cicuit analysis.

From my reasoning, the explaination for this is:
1) The purpose of the short circuit analysis is to calculate the maximum fault current, thus motors are assumed to have a 100% LF (ie, fully energized.)

2) In most cases only the motor's rotor contributes to interia/momentum. Generally speaking, a typical load does not have significant inertia that drives the motor into a generatative state. Typical non momentium loads are pumps and fans. (Regenerative systems such as conveyors going down hill, etc are an exception.)

Am I correct? I'm in learning mode, and always interested in reading your comments.

Thanks :)
 
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The short circuit contribution by a motor is largely independent of the actual mechanical load on the motor at the time of the fault. If the motor is running at speed, its contribution will be basically the same, regardless of load.
 
I realized that DPC, (you just summed up my observation as I am studing short circuit analysis) however, I was heading in the direction of WHY?

I'm just curious on the reasoning why the mechanical load on the motor does not contribute to short circuit? I gave an explaination conducted from my reasoning (I pondered it over lunch), I just want to know if I am right or wrong?

Granted DPC, I'm not trying to sound sour. Always a million thanks for taking the time to write back. :)




 
Look at it the other way - try to explain why it would make a difference? The load on the motor represents power leaving the motor in mechanical form. This results in more electrical power flowing into the motor. But for a fault, the motor becomes a generator and the only source of mechanical energy is rotating mechanical energy in the motor and driven equipment and the stored energy in the magnetic field, unless there is an outside source of excitation. The energy that was going into blowing air or running the conveyor, or whatever, is generally not available to reverse itself back into the fault. And even if it was, the motor contribution would normally be long over since it only lasts a few cycles.

Fault contributions from a generator work the same - the fault contribution from a generator is not a function of the actual load on the generator at the time of the fault.
 
Ah yes, I didn't consider the necessary outside source of excitation for an induction motor. Thanks for pointing me in the right direction.

When the voltage falls to 0V, it will regenerate until its magnetizing flux decays. In motor theory: the magnetizing current establishes the magnetic field so that the motor can spin. Magnetizing current is constant and the amount of magnetizing current that a motor draws depends on how the motor was made and does not vary with load.
In case anyone reading this needs a visual illustration:


In addition, I did take some time this morning to review my reference material, in particular I enjoyed the explaination from HANDBOOK OF ELECTRICAL ENGINEERING by Alan L on pg 294:
"Induction motors react as sub-transient generators during the fault. The magnitude of the subtransient
current is normally taken as the starting current or, more specifically, determined by the air-gap emf and the sub-transient impedance of the induction motor. Since
the induction motor has no external excitation system to create flux, then during a disturbance the flux in the machine is that which is ‘trapped’ in it."

Thanks again DPC :)
 
the load on the motor would change the length of time that the contribution would persist (due to it's inertia) not the magnitude.
 
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