kAIC Calculation

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Carultch

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Massachusetts
reyweesor said:
Yes. But, isn't it the kAIC should be involve in making the schedule of loads?
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KAIC (kilo-amps interrupting capacity) is what it is called, when it is a rating on a breaker.
SCCR (short circuit current rating) is what it is called, when it is a rating on a fuse, a disconnect, or a fused disconnect.

Both of the above may also be called the interrupt rating of an overcurrent device, and refer to the maximum current that the device is rated to safely withstand and trip as intended. Significantly higher than that, and you can expect catastrophic failure. In other words, given current on the circuit between the trip rating and the interrupt rating, the fuse or breaker will do its job. Far above the interrupt rating, the entire device may melt/burn, and fill the box with soot.

Available fault current is what it is called, when you calculate it from site conditions at each point within the circuit.

The KAIC and SCCR ratings of equipment throughout the system has to meet or exceed the available fault current at each location. That is, unless you can take credit for series ratings, which means tested/listed combinations of products, that allow the first device from the source to protect downstream devices. It has to do with the timing of how the first device trips.

You calculate available fault current either by getting the figure from the utility at the service point, or you can get a worst-case value from the specs of the transformer. You need to know the KVA, the impedance (%Z), the secondary voltage, and 1-phase vs 3-phase, to get the worst-case value of secondary-side available fault current. When assuming infinite fault current on the primary side, the full scale formula for fault current of a transformer reduces to the following:

Amps associated with KVA divided by impedance as a decimal.
Example: 500 kVA transformer with 5.75% impedance at 208V/3ph
Amps associated with KVA = 300000/(208*sqrt(3)) =1389A
1389A/0.0575 = 24.154 kA

A worked example of the full scale formula for transformer fault current is available here. Useful if you have fault current across a customer-owned transformer on your site.

Fault current calculations for transformers with Sub-600V on both sides

I'm familiar with calculating KAIC from the fault current calculation spreadsheet available on this website. However, I'm surprised to notice no intake for the voltage on the primary side of the transformer. I know this is geared toward applications where this would be medium voltage and...
forums.mikeholt.com

You can also take credit for fault current decreasing, due to circuit lengths. The impedance of the wire itself reduces fault current with distance from the source. I'm not familiar with the formula for it, as I use a spreadsheet Mike used to host on this website. There are resources available for fault current decrease through conductors, that detail the calculation.

The utility supply is generally the most significant source of fault current, but keep in mind that motors also add to it. A spinning motor that encounters a fault, will effectively become a generator and feed that fault. There are equations that determine fault current contribution from a motor load, based on horsepower and full load current.
 
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topgone

Senior Member
reyweesor said:
Until now i cannot find how to calculate the kAIC rating for circuit breakers and main circuit breakers. Please give me some tips and advice. Thank you.
Click to expand...
Be careful when choosing breakers for the installation point! There are a lot of terms you need to watch out for. If the symmetrical short-circuit current available I at the fault location is known (this maybe what you are asking for all along), the circuit breaker should have a "kAIc" greater the the short-circuit current available at that location! But manufacturers print so many data on the breaker nameplate: Icu =ultimate symmetrical current; Icm =short-circuit making capacity; Ics= rated service short-circuit current, etc. etc. My advise to you is to make sure the available symmetrical short-circuit current available at the location is always lesser than the Ics rating of your chosen circuit breaker!
 
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paulengr

Senior Member
Carultch said:
KAIC (kilo-amps interrupting capacity) is what it is called, when it is a rating on a breaker.
SCCR (short circuit current rating) is what it is called, when it is a rating on a fuse, a disconnect, or a fused disconnect.

Both of the above may also be called the interrupt rating of an overcurrent device, and refer to the maximum current that the device is rated to safely withstand and trip as intended. Significantly higher than that, and you can expect catastrophic failure. In other words, given current on the circuit between the trip rating and the interrupt rating, the fuse or breaker will do its job. Far above the interrupt rating, the entire device may melt/burn, and fill the box with soot.

Available fault current is what it is called, when you calculate it from site conditions at each point within the circuit.

The KAIC and SCCR ratings of equipment throughout the system has to meet or exceed the available fault current at each location. That is, unless you can take credit for series ratings, which means tested/listed combinations of products, that allow the first device from the source to protect downstream devices. It has to do with the timing of how the first device trips.

You calculate available fault current either by getting the figure from the utility at the service point, or you can get a worst-case value from the specs of the transformer. You need to know the KVA, the impedance (%Z), the secondary voltage, and 1-phase vs 3-phase, to get the worst-case value of secondary-side available fault current. When assuming infinite fault current on the primary side, the full scale formula for fault current of a transformer reduces to the following:

Amps associated with KVA divided by impedance as a decimal.
Example: 500 kVA transformer with 5.75% impedance at 208V/3ph
Amps associated with KVA = 300000/(208*sqrt(3)) =1389A
1389A/0.0575 = 24.154 kA

A worked example of the full scale formula for transformer fault current is available here. Useful if you have fault current across a customer-owned transformer on your site.

Fault current calculations for transformers with Sub-600V on both sides

I'm familiar with calculating KAIC from the fault current calculation spreadsheet available on this website. However, I'm surprised to notice no intake for the voltage on the primary side of the transformer. I know this is geared toward applications where this would be medium voltage and...
forums.mikeholt.com

You can also take credit for fault current decreasing, due to circuit lengths. The impedance of the wire itself reduces fault current with distance from the source. I'm not familiar with the formula for it, as I use a spreadsheet Mike used to host on this website. There are resources available for fault current decrease through conductors, that detail the calculation.

The utility supply is generally the most significant source of fault current, but keep in mind that motors also add to it. A spinning motor that encounters a fault, will effectively become a generator and feed that fault. There are equations that determine fault current contribution from a motor load, based on horsepower and full load current.
Click to expand...

Not quite. SCCR is a short circuit withstand rating. AIC is the interrupting rating. The two are related in that they deal with high current faults like ground faults, arcing faults, and shorts but they are definitely not the same. It is common for instance that disconnects and contactors have a much higher SCCR than AIC, especially the non-load break kind. Breakers and fuses are more of an exception than a rule that AIC frequently equals SCCR. So opening one under load AIC will destroy it but another device such as a fuse or load break device does the trick. In medium voltage starters it is common to open the contactor for almost any reason, both overload and other types of faults, but let the fuse handle shorts by preventing opening above AIC. This saves fuses but protects the contactor.
 
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paulengr

Senior Member
topgone said:
Be careful when choosing breakers for the installation point! There are a lot of terms you need to watch out for. If the symmetrical short-circuit current available I at the fault location is known (this maybe what you are asking for all along), the circuit breaker should have a "kAIc" greater the the short-circuit current available at that location! But manufacturers print so many data on the breaker nameplate: Icu =ultimate symmetrical current; Icm =short-circuit making capacity; Ics= rated service short-circuit current, etc. etc. My advise to you is to make sure the available symmetrical short-circuit current available at the location is always lesser than the Ics rating of your chosen circuit breaker!
Click to expand...

That’s just because if you have say 3 engineers you get 6 answers and 1 can argue 2 points with themselves. More data does not mean better answers. It’s just more arguments. Do we really want to get into the myriad of standards for calculating fault current depending on which standard (and assumptions) you want to apply? Even ANSI has two methods, one where we use X and R and another where we just ignore R so slide rules are easier. IEC has probably 6 methods and I’m not even mentioning IEEE.
 
C

Carultch

Senior Member
Location
Massachusetts
paulengr said:
Not quite. SCCR is a short circuit withstand rating. AIC is the interrupting rating. The two are related in that they deal with high current faults like ground faults, arcing faults, and shorts but they are definitely not the same.
Click to expand...

In any case, they are both equipment ratings that have to meet or exceed the available fault current, unless series rated combinations allow otherwise.
 
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