How can fault current be calculated?

[mra65]

how can the fault current be calculated so that the proper utility disconnect can be sized?

 

[shamsdebout]

One possible way is by getting the required information (parameters) and performing a hand calculation as described here: http://www.cooperindustries.com/con...ch_Lib_Short_Circuit_Current_Calculations.pdf

You can also use a calculator. I think there free calculator such as this one: http://www1.cooperbussmann.com/7/FreeSoftware.html

I hope I answered your question correctly.

 

[don_resqcapt19]

You need to start with the available fault current from the utility. Typically they will give you the fault current at the secondary terminals of thier transformer. You can use the calculations in post 2 to get the fault current at the line side of the service disconnect.

 

[shamsdebout]

You need to start with the available fault current from the utility. Typically they will give you the fault current at the secondary terminals of thier transformer. You can use the calculations in post 2 to get the fault current at the line side of the service disconnect.

Thanks for mentioning the starting point :thumbsup:.

 

[BJ Conner]

Short Circuit Calculations

Short Circuit Calculations

how can the fault current be calculated so that the proper utility disconnect can be sized?

Here's the long hard way. Most people now use computer programs like EZpower, SKM etc.

http://www.geindustrial.com/publibrary/checkout/GET-3550F?TNR=White Papers|GET-3550F|generic

 

[G._S._Ohm]

It would be interesting to see how these various calculations match with this measurement.


For 240v, if you have a 240v load that pulls 20A or so, you measure the 240 before and after turning off the load.
If the 240v increases by 0.5 volt when the load is switched off this source is capable of delivering 240v/(0.5v/20A) = 9.6 kA.