Shamp

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charlie b

Moderator
Staff member
Location
Lockport, IL
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Retired Electrical Engineer
Simple: Ohm's Law. Current equals voltage divided by resistance.

But the tricky part, in performing a fault current calculation, is determining the value of resistance that will be seen by the voltage source. Also, if the facility has one or more large motors, the calculation gets trickier, since they will contribute current to the fault point.

Because of these considerations, this type of calculation is generally performed by an engineer or by the distribution equipment vendor, using a special software package. What is it that you are trying to do here?

Welcome to the forum.
 
Fault Currant Calculations

Fault Currant Calculations

The Inspector Wrote a corrections stating that we needed to "verify available fault currant from utility trans"
At the second lug there is 38,000amps
We used Mikeholtz formula but the inspector wants us to do the calculations long hand to further the amps down. So I was hoping that there was a Fault Currant Calculator somewhere that would tell us how to do it long hand...

MarkShamp
 

bphgravity

Senior Member
Location
Florida
Do yourself a favor and hire an engineer. That, or simply provide AIC ratings that equal or exceed that of the AFC at the utility transformer. That way, there is no possibility you are not in compliance.
 

zog

Senior Member
Location
Charlotte, NC
The Inspector Wrote a corrections stating that we needed to "verify available fault currant from utility trans"
At the second lug there is 38,000amps
We used Mikeholtz formula but the inspector wants us to do the calculations long hand to further the amps down. So I was hoping that there was a Fault Currant Calculator somewhere that would tell us how to do it long hand...

MarkShamp

You can use the point to point method if you have a starting point (As you said you do), it is a very long and drawn out method but if you want I will post the procedure.
 

bcorbin

Senior Member
Call the utility to find out what the available fault current is on the primary side of the transformer, then divide by the transformer impedance, which the utility should also have.
 

zog

Senior Member
Location
Charlotte, NC

Ohm's Law: E = IR. You are suggesting dividing I by R. What is that intended to give, as a result? :confused:

Actually the transfomer impedence will be a %, so he is saying divide I by %, dont know what those units would be.

He is confused me thinks, shamp you already have the Isc at the secondary of the transofmrer correct? All you need to do is use the point to point method to show Isc values downstream, if you like I will post the procedure so you can do this by hand to show to your inspector (Whom I am guessing couldnt do the calc by hand himself)
 

jim dungar

Moderator
Staff member
Location
Wisconsin
Occupation
PE (Retired) - Power Systems
Actually the transfomer impedence will be a %, so he is saying divide I by %, dont know what those units would be.

Actually what we call usually call transformer impedance should be written as %IZ therefore the units could be called volts. But because % is similar to the PU it is basically unit-less.
 

bcorbin

Senior Member
Oops...my fingers must have been working faster than my brain. What I really meant was use the utility figure or divide the rated current for the transformer by the impedance, as a quick rule of thumb.
 
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