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Max Fault Current or Amperage Interrupt Capacity(AIC) calculation

rjeezy (Electrical) (OP)
16 Jun 11 12:59
How to calculate Max Fault Current or Amperage Interrupt Capacity?


Given that it's a three phase load, consisting of 2 fuses(40 Amps, 200kA AIC), 2 Variable Fixed Drives(15 HP, 10 kA AIC) and two contacters( 5 kA AIC). I'm interested in finding the Max Fault Current or also known as the the total Amperage Interrupt Capacity(AIC). From looking online, I know that its (rated current/ &impedance)*100. Will I able to determine the AIC with these given data, or do I need more?
Zogzog (Electrical)
16 Jun 11 13:12
AIC is a device rating, it is the amount of fault current a device is capable of interupting, not something you can calculate.

Available fault current is something you can calculate and is dependent on the source feeding the circuit, you also must account for motor contribution. There are several ways to do this calculation but not a simple forumla.

At each point in your system the available fault current may not exceed the AIC rating of the device.  
waross (Electrical)
16 Jun 11 13:16
%impedance based calculations yield a value known as "Available Short Circuit Current."
This is also called the symmetrical short circuit current. These ASCC
values are used to select equipment with adequate fault interrupting capacity but they are not the total possible short circuit current.  
Many fault currents are initially asymetrical. DC offset is a term that is often used to describe the onset of a fault. The amount of dc offset, or the actual  peak current will depend on the X/R ratio of the source and the point on the waveform that the fault occurs.
The actual interupting capacity of a breaker rated at 10 kAIC will be a value that corresponds to the maximum peak current possible based on a representative value of the X/R ratio of the source.
Use the %impedance of the supply transformer to select equipment that will safely interrupt the actual possible fault currents.

Bill
--------------------
"Why not the best?"
Jimmy Carter

ThePunisher (Electrical)
17 Jun 11 0:22
You may take the supply transformer kVA rating and divide it with it's rated % impedance based on its KVA rating and assume an infinite source. If you do not have any large motor contribution, this will be conservative enough.

Better yet, you may calculate the symmetrical short circuit based on an MVA base and equivalent P.U. impedance at the fault point. If the calculated X/R at fault does not exceed the fuse or breaker X/R test, then you may directly compare the calculated symmetrical current against the KAIC rating.  
waross (Electrical)
17 Jun 11 10:58
What ThePunisher said, but be aware that the ratings and testing methods take into account that the actual current may be much greater than calculated. The use of "Available Short Circuit Currents" makes it easy to select and match equipment.  

Bill
--------------------
"Why not the best?"
Jimmy Carter

jraef (Electrical)
17 Jun 11 20:09
rjeezy,
No offense, but it appears that you are in over your head here. You are not sure of the terms to use, let alone why this is important or how it all works. You are describing the process of determining the SCCR of a control system, not the AFC or the AIC. It's a very involved process that cannot be done justice to in a limited format such as this and it is definitely not something for amateurs to tackle.

You need the services of a qualified Electrical Engineer for this. If you are an EE, this is obviously outside of your level of expertise. We cannot substitute for that over the internet on something so involved. Please seek out the proper assistance from someone who can observe your entire system.

"Dear future generations: Please accept our apologies. We were rolling drunk on petroleum."
— Kilgore Trout (via Kurt Vonnegut)
  
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