Calculating available fault current

[Jpflex]

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

 

[ggunn]

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.

 

[HEYDOG]

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.

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.

Here is a link to short circuit calculations. https://www.eaton.com/content/dam/e...nter/bus-ele-tech-lib-electrical-formulas.pdf

 

[Jpflex]

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?

 

[augie47]

Eaton also has a fault curent app... just plug in your numbers....

Fault Current Calculator - Apps on Google Play

A tool to calculate & print available fault current on electrical systems.

play.google.com

 

[mayanees]

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

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.