Fault current

[paulnorris]

How do I figure available fault current?

 

[charlie b]

Available where (e.g., at a building's service point or at a distribution panel within the building), and from what source (e.g., from a utility transformer, an on-site transformer, or a portable generator)?

 

[mivey]

How do I figure available fault current?

Based on what information? If you have a 3-phase transformer with impedance "Z", the available fault at the transformer will be:
kVA / (sqrt(3) * L-L kV * Z)

This is a conservative number as this assumes an infinite source bus. (The source impedance would reduce this number)

[edit: clarified formula]

 

[paulnorris]

Thanks. I want to figure available fault current at the transformer. Example would be 45 KVA 277/480 3 ph 3.01 imp.

 

[augie47]

waht Chalie b says..

but, pardon me, looking at your profile I can;'t help but say WHATTTTT ?????
we always call YOU

 

[mivey]

Thanks. I want to figure available fault current at the transformer. Example would be 45 KVA 277/480 3 ph 3.01 imp.

Then it would be 45 / (sqrt(3)* 0.480 * 0.0301) = 1798 amps

[edit: I used 0.0301 because Z is usually given in % so 3.01% = 0.0301]

 

[paulnorris]

Thanks Mivey.

 

[cpal]

Im surprised the power company doesn't have namograms or a computer program to perfom this calculation. ????

 

[paulnorris]

It's something I want to be clear on if I figure it out on paper in the field.

 

[GMIEE]

While this infinite bus calculation is great, it is of no use for arc flash analysis am I correct? Just checking.

 

[mivey]

While this infinite bus calculation is great, it is of no use for arc flash analysis am I correct? Just checking.

Correct, you need more info for the arc flash calculations.

 

[mivey]

Paul, also see posts #9 #10, #13 on

http://forums.mikeholt.com/showthread.php?t=96566