Circuit Breaker kAIC ratings
- Thread starter jimmac49
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ron
Senior Member
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- New York, 40.7514,-73.9925
Re: Circuit Breaker kAIC ratings
The rating of the breaker by the manufacturer itself begins with an educated guess. They bring it to a test lab and see if it complies with the UL testing results. If it doesn't, then they try again with a lower rating. If it can't pass tests at 10kAIC, then it will not come to market, since the minimum rating generally available is 10kAIC.
The rating of the breaker by the manufacturer itself begins with an educated guess. They bring it to a test lab and see if it complies with the UL testing results. If it doesn't, then they try again with a lower rating. If it can't pass tests at 10kAIC, then it will not come to market, since the minimum rating generally available is 10kAIC.
- Location
- Lockport, IL
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- Retired Electrical Engineer
Re: Circuit Breaker kAIC ratings
Was your question intended to address the calculation of the system's "available fault current," or the rating of the breaker? It sounds like the later, and to that question I will second Ron's answer.
But if you are asking about the available fault current, that is a "simple" application of Ohm's Law. The "I" is found by simply dividing the "V" by the "R." However, it is a complicated matter to obtain a model of the system, in order to get valid values for "V" and "R."
Was your question intended to address the calculation of the system's "available fault current," or the rating of the breaker? It sounds like the later, and to that question I will second Ron's answer.
But if you are asking about the available fault current, that is a "simple" application of Ohm's Law. The "I" is found by simply dividing the "V" by the "R." However, it is a complicated matter to obtain a model of the system, in order to get valid values for "V" and "R."
BodnarMatt
Member
Re: Circuit Breaker kAIC ratings
Generally one must first calculate the maximum fault current for the transformer secondary using the equation:
I(maxSC) = (XFRMR KVA x 100)*(100% - Trasformer impedance) / (Secondary L-L Voltage x 1.732)
Then calculate the point-to-pont short circuit current:
I(SC) = I(maxSC) * (1 / (1 + (1.73 * ft * I(maxSC))/ (Const. * L-L voltage)
where the constant is determined by the rating on the busway 225 A -> Const. = 28700
400A -> Const. = 38900
600A -> Const. = 41,100
800A -> Const. = 46,100
etc.
This would give the short circuit current a number of feet from the the transformer at the bus plug.
If there are large loads that tend to be regenerative motors they can add to the fault current. One would usually use some software to determine the fault at a panel, it would take into account all wire size and length, etc.
Generally one must first calculate the maximum fault current for the transformer secondary using the equation:
I(maxSC) = (XFRMR KVA x 100)*(100% - Trasformer impedance) / (Secondary L-L Voltage x 1.732)
Then calculate the point-to-pont short circuit current:
I(SC) = I(maxSC) * (1 / (1 + (1.73 * ft * I(maxSC))/ (Const. * L-L voltage)
where the constant is determined by the rating on the busway 225 A -> Const. = 28700
400A -> Const. = 38900
600A -> Const. = 41,100
800A -> Const. = 46,100
etc.
This would give the short circuit current a number of feet from the the transformer at the bus plug.
If there are large loads that tend to be regenerative motors they can add to the fault current. One would usually use some software to determine the fault at a panel, it would take into account all wire size and length, etc.
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