Are you calculating asymetrical system fault currents for the purpose of protection and co-ordination, or are you looking for available symetrical currents for the application of symetrical rated circuit breakers.
For the former you will probably use computers and simulation software.
If you wish to determine if the equipment in a small installation will withstand the available fault current from a local distribution transformer, divide the rated secondary current of the transformer by the percent impedance voltage of the transformer to find available symetrical fault current. If the available fault current is a little over the rating of the installed devices, calculate the impedance of the supply conductors to see if this will limit the current to an acceptable limit.
The circuit breakers are rated in symetrical current and the rating includes a safety factor to allow for the higher asymetrical fault currents.
The challenge is to be able to recognise when a system is simple enough to use the simple calculations and when the system complexity requires a more detailed analysis.
yours
Ohm Method
SC Current = (Voltage before SC)/ (System Impedance @ fault location).
Isc =V/Z = V(R+jX) A variation of this method, is the MVA or KVA method.
Per Unit (PU) Method
Often the calculation uses the PU method to normalize values with an equivalent circuit often of 1 pu voltage or close to this value.
In this case:
SC Current=1/(System Impedance @ fault location) = Admittance in PU.
waross,
the utility company came back to me with a 3ph bolted
fault of 13,727 amperes (r.m.s.)(7,500 +j5100 microhms)
that is greater than the aic of the mdp installed (and of course all the downstream panels)
i have to change out the cb's, label the panels per nec and provide short circuit calculations. Is this correct?
thank you for your help.
Why on earth would electrical equipment have been ordered and installed before getting the available fault current from the utility. Sounds like getting the cart before the horse.
Are you fed directly off the utility? Was the S/C value for the primary or seconday of your transformer? Typically the utility gives you a starting point, the S/C current will go down from there.
I think you got the point from other posters that S/C calcs are not that simple, you are going to want to get a S/C study, then see if you have a problem.
Here is a simplified and quick method to verify if the AIC of the CB rating is OK:
a) Obtain the impedance of the feeder from the utility to CB
b)If the utility SC available is before the distribution transformer, determine the transformer impedance.
c) Add the utility impedance to the above impedances, a + b and obtain magnitude of the resulting complex number=|Z|.
The available SC at the CB panel is approximately the ratio of the nominal voltage divided by |Z|.
It is my understanding that the available short circuit current calculated with the percent impedance voltage of the transformer is based on an infinite supply. As a result it is a conservative figure.
If the 13,727 amps is at the transformer secondary, there may well be enough impedance in the conductors to the main panel to get below 10,000 amps, if that helps.
If the 13,727 amps is at the panel, that's the figure that you have to work with.
Another option may be to add series impedance to the feeders to limit the current. I don't know if they are still on the market, but at one time wireless reactors were available. This was a bracket and a stack of cores on the front of the transformer through which the feeders passed. They could add several percent impedance to a circuit. Be aware that as you reduce the available bolted fault current, the voltage regulation will suffer in the ratio of the change. You can probably accept the change.
respectfully
waross
In most cases here it doesn't matter what the real current is going to be. The inspector will want to see what the utility gives as the availabel fault current.
The utility may install a transformer with a 5% impedance and caculate the fault current base on a 2% impedance transformer. The theory being that somewhere in a storage yard they have a 2% impedance transformer that may get installed at that location someday.
Hello BJC
I don't have a problem with your post. In fact I consider it to be pretty practical in the real world.
My point was that a small amount of reactance can be shown to reduce the available current at the circuit breaker to a value less than the bolted fault current.
respectfully
The figure provided by the utility is bigger than your main incomer's rating - nothing you can do about it - you have either to limit the short circuit current on a way or upgrade your breaker. Regarding your downstream panels - the ratings might be sufficient if you have a big enough impedance between them and the incomer, but it might also be to small. Hand calculations are rather conservative - in this case I would find someone who can run a computer program to estimate the different current-levels. If you are lucky you need not to upgrade your downstream system.
Regarding fault limiting reactors - we bought some 11kV reactors a few years ago - and I believe it will still be manufactured. What is your system voltage?
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The lower the voltage, the quicker the fault current is reduced by conductor impedance, so your downstream panels may be ok, good calculations are needed. There was a thread a while back that was essentially a rant about the fault current values provided by the utility companies and how they can have the effect of increasing the required rating of the customer's equipment. The end result is that while the utility numbers are likely higher than will ever be seen, they are the only numbers that the local inspectors will accept and therefore they are the numbers that must be used. The point of the numbers, from the utility's perspective, is that they will give you a number that they can never exceed; so, of course it will be higher than you will probably ever see. Consider it a built in safety margin.
Is it common for the utility to calculate feeder impedance and report fault levels at the main breaker or is the reported fault level at the transformer terminals.
I understand that it is probably too late here, but I have seen fault levels limited by running the 13kv primary feeders in separate 2" rigid conduits for several hundred feet.
I have seen secondary fault currents limited by specifying a minimum feeder length for each unit sub. The extra cable was run past the destination and then doubled back.
The secondary reactors were cores through which the secondary cables were passed.
Primary reactors will work as well as secondary reactors.
stevedantonio;
Can you give us a little information on the system?
KVA ratings? Is your main panel bus bar connected to the transformer or connected by feeders. Are the feeders long or short?
Can you limit the current by the installation of a length of cable on either the primary or secondary feeders, and will it be any cheaper than changing the main breaker. Can you add enough cable for each of the sub feeds to avoid having to change all of the panel breakers and will the cable be cheaper than the breakers?
I think your options are either change the breakers, or reduce the fault current with reactance. This can be either reactors on the primary or secondary or suitable lengths of cable. From there it is an economic choice as to which is the most economical. Be aware that the additional reactance will have an small effect on your voltage regulation.
respectfully
There is also the small problem of any type of impedance used to reduce fault current will also produce a standing voltage drop; another example of no free lunches.
The short circuit current provided by the utility is 13.7kA. What is your short-circuit rating/s of your breakers? If the difference is not too much, and you can accept some volt-drop, you might get away using some extra cable. Certainly worth consideration - maybe an added volt-drop resulting in no breaker changes.
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To illustrate how rapidly the SC is reduced in low voltage applications, enclose is a link that shows an example for 225 kVA transformer with various feeder size as function of the cable length.
Notice that a SC of 13K to 15 kA at the transformer bushing will be reduced to 10 kA at 20 ft and 5 kA for 60 ft for various utility SC ranging from infinite to 50,000kVA.
