Calculating available fault current

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Jpflex

Jpflex

Electrician big leagues
Location
Victorville
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Electrician commercial and residential
In the process of labeling disconnects and service equipment for my company at a mine, I noticed no label for available fault current.

Some equations online say to get KVA from transformer and divide by voltage X (1.732 if 3 phase) X Z (impedance)

Others say to Omit z out of equation above use Z as a multiplier to KVA/ volts x 1.732

Z = 100/(Z x plus or minus percent or 0.9)

Therefore the way mike holt showed this is different from the last method

Can I just multiply divisor bottom fraction by impedance shown on transformer name plate? Additionally I’m working on single phase so i won’t be factoring in 1.732. Nor do I know how to factor in service or feeder impedance yet

What is the easiest formula? Thanks
 
Using the infinite bus model on the transformer primary:

Available Fault Current (AFC) in kA at a three phase utility transformer secondary = (kVA)(100) / (sqrt(3))(V line to line)(%Z)

I have verified this using data from a chart of transformer kVA, %Z, and AFC given me by my local utility.
 
ggunn said:
Using the infinite bus model on the transformer primary:

Available Fault Current (AFC) in kA at a three phase utility transformer secondary = (kVA)(100) / (sqrt(3))(V line to line)(%Z)

I have verified this using data from a chart of transformer kVA, %Z, and AFC given me by my local utility.
Click to expand...
Jpflex said:
In the process of labeling disconnects and service equipment for my company at a mine, I noticed no label for available fault current.

Some equations online say to get KVA from transformer and divide by voltage X (1.732 if 3 phase) X Z (impedance)

Others say to Omit z out of equation above use Z as a multiplier to KVA/ volts x 1.732

Z = 100/(Z x plus or minus percent or 0.9)

Therefore the way mike holt showed this is different from the last method

Can I just multiply divisor bottom fraction by impedance shown on transformer name plate? Additionally I’m working on single phase so i won’t be factoring in 1.732. Nor do I know how to factor in service or feeder impedance yet

What is the easiest formula? Thanks
Click to expand...
ggunn said:
Using the infinite bus model on the transformer primary:

Available Fault Current (AFC) in kA at a three phase utility transformer secondary = (kVA)(100) / (sqrt(3))(V line to line)(%Z)

I have verified this using data from a chart of transformer kVA, %Z, and AFC given me by my local utility.
Click to expand...
Here is a link to short circuit calculations. https://www.eaton.com/content/dam/e...nter/bus-ele-tech-lib-electrical-formulas.pdf
 
Also I do go off final step down transformer feeding breaker panel Prior to making calculations but not those Upstream closer to utility but still on company site?
 
Jpflex said:
In the process of labeling disconnects and service equipment for my company at a mine, I noticed no label for available fault current.

Some equations online say to get KVA from transformer and divide by voltage X (1.732 if 3 phase) X Z (impedance)

Others say to Omit z out of equation above use Z as a multiplier to KVA/ volts x 1.732

Z = 100/(Z x plus or minus percent or 0.9)

Therefore the way mike holt showed this is different from the last method

Can I just multiply divisor bottom fraction by impedance shown on transformer name plate? Additionally I’m working on single phase so i won’t be factoring in 1.732. Nor do I know how to factor in service or feeder impedance yet

What is the easiest formula? Thanks
Click to expand...
Make sure your label says maximum available fault current since you're using the infinite primary method. Available fault current would require a Power Study that gets contribution information from the Utility for the fault current analysis.
 
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