?Short Circuit Current Rating?

[Russell Ensslin]

What's a Short Circuit Current Rating? (SCCR)
Definition says, "Prospective symmetrical fault current...able to be connected..."
What is the "Prospective" part?
What is the "Symmetrical" part?
Why is the incoming current assumed to be "Fault Current" if it is the device receiving the current?
It is referring to current being received only, isn't it?
So if I have a GFCI receptacle with a SCCR of 10,000 amp being protected by a 20 amp, over-current device with a 10,000 amp SCCR protected by a 200 amp main with a SCCR of who-knows-what, then some current would have to blow through the main breaker and then through the 30 amp breaker to send 10,000 amps to the GFCI?
Is that what this is saying?
Wouldn't the main breaker trip after 200 amps? And if it failed, wouldn't the 30 amp breaker trip after only 30 amps? Is this all just theoretical? Or, perhaps, in the event of lightning or something like that?

 

[romex jockey]

Russ,
basically it's the momentary blast of energy any given electrical component can withstand

~RJ~

 

[GoldDigger]

There will always be a finite time before a breaker can trip. Under some conditions (small overcurrent) that time can be minutes. For short circuit faults the time will be relatively short, often less than 1/2 cycle depending on when in the cycle the fault occurs.
So the SCCR of a device is related to how much current can safely pass through that device during the time before the OCPD trips.
Some types of OCPD (like fuses) may guarantee a maximum value that the current can reach during the trip event. The downstream device SCCR must be greater than that.
In the case of a typical breaker, the breaker does not enforce that kind of limit and the current the downstream device has to tolerate for a short time will be based on the available fault current from the power source to the breaker. That is an area where symmetric and asymmetric fault currents come into play.

The assumption is that the fault is somewhere downstream of the device (like a switch or whatever) or inside the device (like an appliance) caused by an internal failure. An internal fault will dissipate energy inside the device too, and the amount that it can tolerate depends on the current available from upstream.

 

[don_resqcapt19]

The SCCR is the maximum fault current that the device can handle without damage. Often this rating has a time associated with it. This rating is most often used with passive equipment, that is equipment that is not designed to interrupt the fault current. The overcurrent devices will have an AIC (ampere interrupting current). This is the maximum fault current that the OCPD is designed to interrupt without damage.

The available fault current is the maximum current that the system can supply under fault conditions. The OCPDs must have an AIC that exceeds the available fault current and the other equipment must have a SCCR that exceeds the available fault current.

The available fault current goes down rapidly as you move away from the source because of the impedance of the system. For example if you have 20,000 amps of available fault current at the load side of a 200 amp breaker and you install 50' of 4/0 copper in a steel conduit, the available fault current at the load end of those conductors is 16,140 amps. If you would increase the run to 150' the available fault current would go down to 11,644.

In your example if the GFCI is 50' away from the source with 10,000 amps of available fault current(line to line fault current) and you use #12 NM, between the source and the GFCI you would have 705 amps of line to neutral fault current available at the GFCI

 

[augie47]

You can find some interesting articles on AIC, SCCR etc. at the gear manufacturers sites.
I did have a couple of reference that were a bit better than this one, but this is pretty good:
http://www.cooperindustries.com/con...ch_Lib_Short_Circuit_Current_Calculations.pdf

 

[kwired]

The amount of current that can flow when a short circuit happens depends on the impedance of the source. Smaller sources in general have a higher impedance. The resistance of conductors between the source and the point of interest in determining fault current levels also limits how much current will flow during a fault.

Lets say that 30,000 amps is capable of flowing because of the components used, if you have a fault current will rise to that level, it takes the overcurrent device time to respond to it - may only be a few milliseconds but high current does flow for the duration it takes until the circuit is opened.

If a device is only rated to withstand 10,000 amps (SCCR or AIC whichever applies) it can be damaged if the available current is higher then the rating. This damage can be violent in nature as well.

Most(probably all) of your general use type circuit breakers are rated 10,000 amps. Anything not marked is usually considered to be able to handle 5000 amps.

Class RK5 (most common time delay fuses other then plug type fuses) fuses are typically rated 100,000 amps. There are other breakers rated more then 10,000 - most often they are mains or other feeder breakers and not your typical general use branch breakers.

We then get into more complicated applictions where series ratings come into play where an item with a lower rating can be used if a specific item it has been series tested with is ahead of it in the circuit.

 

[Russell Ensslin]

Russ,
basically it's the momentary blast of energy any given electrical component can withstand

~RJ~

This was useful - Thanks

 

[Russell Ensslin]

The SCCR is the maximum fault current that the device can handle without damage. Often this rating has a time associated with it. This rating is most often used with passive equipment, that is equipment that is not designed to interrupt the fault current. The overcurrent devices will have an AIC (ampere interrupting current). This is the maximum fault current that the OCPD is designed to interrupt without damage.

The available fault current is the maximum current that the system can supply under fault conditions. The OCPDs must have an AIC that exceeds the available fault current and the other equipment must have a SCCR that exceeds the available fault current.

The available fault current goes down rapidly as you move away from the source because of the impedance of the system. For example if you have 20,000 amps of available fault current at the load side of a 200 amp breaker and you install 50' of 4/0 copper in a steel conduit, the available fault current at the load end of those conductors is 16,140 amps. If you would increase the run to 150' the available fault current would go down to 11,644.

In your example if the GFCI is 50' away from the source with 10,000 amps of available fault current(line to line fault current) and you use #12 NM, between the source and the GFCI you would have 705 amps of line to neutral fault current available at the GFCI

.


Thank you Don. This is helping

So if you shoot a rocket into an electrically charged cloud with a tiny copper wire and lightning strikes, the wire will be vaporized but not until it will have carried the "ground fault" back to earth.
Could you call the SCCR the amount (percentage of the charge of the lightning) of charge the wire could deliver before it lost continuity? Is that the SCCR?
On that same note, can the main OCPD and the 30 amp SCCR be somehow blown through by a blast of current that is delivered before the OCPD can break the circuit?

In other words, is the SCCR the amount of charge that can be stopped by the OCPD before it will be smoaked by the intruding surge?

 

[GoldDigger]

.


Thank you Don. This is helping

So if you shoot a rocket into an electrically charged cloud with a tiny copper wire and lightning strikes, the wire will be vaporized but not until it will have carried the "ground fault" back to earth.
Could you call the SCCR the amount (percentage of the charge of the lightning) of charge the wire could deliver before it lost continuity? Is that the SCCR?
On that same note, can the main OCPD and the 30 amp SCCR be somehow blown through by a blast of current that is delivered before the OCPD can break the circuit?

In other words, is the SCCR the amount of charge that can be stopped by the OCPD before it will be smoaked by the intruding surge?

Not quite. What you describe is the interrupting capacity of an OCPD. An OCPD does not have an SCCR rating. The devices with SCCR ratings are ones that do not interrupt the flow of energy and just have to survive the current passing through it until the OCPD somewhere else opens.

 

[FionaZuppa]

what does this amps short rating mean if you dont include a time factor?

10kA for 0.00000001sec is much different than say 5kA for 0.001sec

a small coulomb discharge (short) from a cap can yield 20kA.

 

[charlie b]

We don't yet have an answer to the question of what symmetrical and assymetrical mean. I thought I knew, but now I am less certain. I tried some on-line research, and found two separate and entirely different answers.

• One answer is that a fault can be called symmetrical if it impacts all three phases in an identical manner. Examples are the L-L-L fault and the L-L-L-G fault (i.e., the three phase bolted fault and the three phase to ground fault). According to this version, a fault is assymetrical if it impacts one of the phases differently than the others. Examples are the single line to ground fault and the line to line fault. I don't think this is the right answer.
• The other answer is that the behavior of the current from moment-to-moment (in terms of how high it gets on one cycle and how much it drops off on the next cycle) will depend on the moment in time that the fault occurs (i.e., whether the voltage is crossing the zero line or is at its positive peak or is at its negative peak or is somewhere in the middle). Regarless of when the fault occurs, there will be a large AC component of current that will look like a basic sine wave but will get smaller and smaller on each cycle. However, depending on when the fault occurs, there may also be a DC component added to the AC component. In other words, current will start higher, and the curve will look like it has been shifted upwards (i.e., the positive peaks are higher above the axis than the negative peaks are below the axis). So if there is no DC component, you have a symmetrical fault, and if there is a DC component, you have an assymetrical fault. Here again, I am no longer whether the DC component will come into play if the fault happens with the voltage at its zero point or if the fault happens when the voltage is at its peak value.

Can someone fill in the blanks for me?

 

[FionaZuppa]

see http://www.egr.unlv.edu/~eebag/Symmetrical Fault Analysis.pdf


i suspect your 2nd notion is not related because under all cases the short current will decrease over time if the source of the short is constant voltage, this due to heating which increases ohms of the metal. if the source is a like a cap then the short current decreases faster because voltage drops off and metal gets warmer during the sort.

 

[jim dungar]

• The other answer is that the behavior of the current from moment-to-moment (in terms of how high it gets on one cycle and how much it drops off on the next cycle) will depend on the moment in time that the fault occurs (i.e., whether the voltage is crossing the zero line or is at its positive peak or is at its negative peak or is somewhere in the middle). Regarless of when the fault occurs, there will be a large AC component of current that will look like a basic sine wave but will get smaller and smaller on each cycle. However, depending on when the fault occurs, there may also be a DC component added to the AC component. In other words, current will start higher, and the curve will look like it has been shifted upwards (i.e., the positive peaks are higher above the axis than the negative peaks are below the axis). So if there is no DC component, you have a symmetrical fault, and if there is a DC component, you have an assymetrical fault. Here again, I am no longer whether the DC component will come into play if the fault happens with the voltage at its zero point or if the fault happens when the voltage is at its peak value.

UL Listed SCCR and AIC values are based on this use of the term symmetrical (I believe they use the phrase steady-state) in relation to fault current 'withstandability' as raised by the OP question. Published ratings do not address this asymmetry except through the use of the X/R ratio of the circuit.

 

[Russell Ensslin]

Not quite. What you describe is the interrupting capacity of an OCPD. An OCPD does not have an SCCR rating. The devices with SCCR ratings are ones that do not interrupt the flow of energy and just have to survive the current passing through it until the OCPD somewhere else opens.

During Mike's Code course, chapter 1, he talks about the Short Circuit Current Rating for equipment and shows a 30 amp breaker and stamped on the side of it, it shows the SCCR amps of 10,000 and said most equipment has this.
My question is how can this be 10,000 if only thirty can go through?
I think the answer is that it can stop all but thirty of the 10k amps, after 10k I think it fries out. Other posts on this thread say SCCR has to do with what's down stream of the OCPD which is confusing

 

[Russell Ensslin]

We don't yet have an answer to the question of what symmetrical and assymetrical mean. I thought I knew, but now I am less certain. I tried some on-line research, and found two separate and entirely different answers.

• One answer is that a fault can be called symmetrical if it impacts all three phases in an identical manner. Examples are the L-L-L fault and the L-L-L-G fault (i.e., the three phase bolted fault and the three phase to ground fault). According to this version, a fault is assymetrical if it impacts one of the phases differently than the others. Examples are the single line to ground fault and the line to line fault. I don't think this is the right answer.
• The other answer is that the behavior of the current from moment-to-moment (in terms of how high it gets on one cycle and how much it drops off on the next cycle) will depend on the moment in time that the fault occurs (i.e., whether the voltage is crossing the zero line or is at its positive peak or is at its negative peak or is somewhere in the middle). Regarless of when the fault occurs, there will be a large AC component of current that will look like a basic sine wave but will get smaller and smaller on each cycle. However, depending on when the fault occurs, there may also be a DC component added to the AC component. In other words, current will start higher, and the curve will look like it has been shifted upwards (i.e., the positive peaks are higher above the axis than the negative peaks are below the axis). So if there is no DC component, you have a symmetrical fault, and if there is a DC component, you have an assymetrical fault. Here again, I am no longer whether the DC component will come into play if the fault happens with the voltage at its zero point or if the fault happens when the voltage is at its peak value.

Can someone fill in the blanks for me?

Thank You Charlie
I'm honored and flattered that you have taken time to research this.
Usually when I'm trying to understand something I ask questions about it's definition.
But in this science I've come across some whacky nomenclature. Maybe that's what this is. Have you any ideas why it is called a "short" circuit?

 

[augie47]

During Mike's Code course, chapter 1, he talks about the Short Circuit Current Rating for equipment and shows a 30 amp breaker and stamped on the side of it, it shows the SCCR amps of 10,000 and said most equipment has this.
My question is how can this be 10,000 if only thirty can go through?
I think the answer is that it can stop all but thirty of the 10k amps, after 10k I think it fries out. Other posts on this thread say SCCR has to do with what's down stream of the OCPD which is confusing

The breaker rating of 30 amps indicates the current level it is designed to allow to flow.
Current in excess of 30 amps will cause the breaker to open. The time it takes the breaker to open is determined by the amount of current istsenses. On a slight overload it may take a matter of minutes, on a short or heavy overload it might open in cycles.
In event of a short (L-L fault, L-N fault, L-G fault) the breaker might well open in cycles. During one of these fault conditions, the amount of current that the breaker actually interrupts is the fault current. It can amount to thousands, tens of thousands or hundreds of thousand of amps dependent on a number of factors. The magnitude of current that breaker can handle safely is indicated by the AIC rating. That is why some 15 amp breakers will fit in your hand and some are as big as a bread box. It has to do with heat dissipation when operating at high current levels.

 

[charlie b]

If you push more current through a breaker than it is rated to handle, it is possible (even likely) that the heat resulting from that much current will start to melt the metal contacts, essentially welding them together. Thereafter, it will not be possible for the breaker to open, and the exact nature of the subsequent destruction is anybody's guess. That is why breakers are given AIC ratings: "Amps Interrupting Capacity." They can interrupt 10,000 amps. Above that, they can no longer interrupt the flow of current, because their contacts will have welded themselves together.

That is also why the term AIC cannot be properly used in conjuction with a panelboard. The panelboard does not have the ability to interrupt the flow of current. Its circuit breakers do that job, not the panelboard itself.

 

[charlie b]

Have you any ideas why it is called a "short" circuit?

The "long road" that current normally takes starts at the panelboard, runs along a wire, goes through an element (i.e., the load, perhaps a light bulb associated with a floor lamp) that possesses the resistance that reduces the amount of current to a manageable level, and finally returns to the panelboard via another wire. If you were to cut the lamp's power cord with a pair of scissors (DO NOT TRY THIS AT HOME!), and if the scissors hit both of the wires at the same time, then the current flow has a shorter path to follow than it once had. It goes from the panelboard via one wire (not as long as it was before), through the blade of the scissors, into the other wire, and immediately back to the panelboard. Since the light bulb (a high resistance element) is now in parallel with the scissors blade, the total resistance of the circuit is much lower. Thus, the total current is much higher, as evidenced by the arcing and sparking at the location of the scissors blade.

 

[FionaZuppa]

The "long road" that current normally takes starts at the panelboard, runs along a wire, goes through an element (i.e., the load, perhaps a light bulb associated with a floor lamp) that possesses the resistance that reduces the amount of current to a manageable level, and finally returns to the panelboard via another wire. If you were to cut the lamp's power cord with a pair of scissors (DO NOT TRY THIS AT HOME!), and if the scissors hit both of the wires at the same time, then the current flow has a shorter path to follow than it once had. It goes from the panelboard via one wire (not as long as it was before), through the blade of the scissors, into the other wire, and immediately back to the panelboard. Since the light bulb (a high resistance element) is now in parallel with the scissors blade, the total resistance of the circuit is much lower. Thus, the total current is much higher, as evidenced by the arcing and sparking at the location of the scissors blade.

hmmm, but what happens if a 1meg ohm carbon resistor decides to become 20ohm? or the filament in that bulb decides to stretch out to where the midpoint is now hanging on the input lead. the path is still the same length, not really shorter


if i recall right, a short in any circuit where the current is more than expected by design. so something to some extent has "shorted", in the sense that the ohms has become "shorter" or "less" or "smaller". so if your motor winding has Z where math and testing says 120v @60Hz says 20A and at some point you start to see 25A, there's likely some type of "short" going on.

certainly a scissor cut of a lamp cord is a type of short.

 

[don_resqcapt19]

A short circuit is a conductive path between circuit conductors other than the connected load(s).