short circuit calculations
short circuit calculations
(OP)
how are these calculations performed?
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short circuit calculations
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short circuit calculationsshort circuit calculations(OP)
how are these calculations performed?
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RE: short circuit calculations
Or you can refer to IEEE "Buff" Book for Recommendations for Power System Protection and Coordination.
or the like.
RE: short circuit calculations
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
RE: short circuit calculations
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.
In PU: Isc =1/Z = Y
RE: short circuit calculations
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.
RE: short circuit calculations
RE: short circuit calculations
getting the cart before the horse is exactly what happened
on this job
RE: short circuit calculations
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.
RE: short circuit calculations
all i know is that the bolted fault mentioned above could be expected at the point-of-service.
thank you
RE: short circuit calculations
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|.
The enclose form could guide you in this matter.
http://ww
If this calculated value is lower than the CB AIC, the rating of the CB is acceptable. Otherwise, upgrade the CB with one of higher AIC.
ADVISE: Get a qualified EE to verify your calc.
RE: short circuit calculations
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
RE: short circuit calculations
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.
RE: short circuit calculations
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
RE: short circuit calculations
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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RE: short circuit calculations
RE: short circuit calculations
RE: short circuit calculations
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
RE: short circuit calculations
RE: short circuit calculations
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.
Failure seldom stops us, it is the fear for failure that stops us - Jack Lemmon
Make the best use of Eng-Tips.com
Read the Site Policies at FAQ731-376
RE: short circuit calculations
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.
ht
RE: short circuit calculations
I love your graphics. Thanks for ilustrating what I was trying to explain.
davidbeach
Have you chosen the wrong word or have I misunderstood you? Adding impedance will cause a voltage drop that is proportional to the current. If a reactance is added in series with a 2% impedance transformer of a value that results in a combined transformer plus reactor impedance of 3% the symetrical short circuit current and the voltage regulation will be the same as it would be with a 3% impedance transformer without reactors.
The phrase "standing voltage drop" may be misleading.
Respectfully
RE: short circuit calculations
I had not addressed voltage regulation, and you are correct that voltage regulation to the load would be the same across 3% worth of impedance whether it is one 3% impedance transformer or a 2% transformer plus a 1% impedance.
My phrase "standing voltage drop" referred to the situation where there is a reasonably steady load and the utility transformer is set for a secondary voltage. If the separate impedance were in the circuit, there would be a voltage drop (probably more of a phase angle shift than drop depending on the X/R of the impedance) across the impedance that wouldn't be there if the impedance wasn't there and that is the standing voltage drop.
RE: short circuit calculations
Thanks for the reply. I think we are in agreement.
As to voltage adjustment, In my experience this is usually done on the basis of open circuit voltage to compensate for cronically high or low primary voltages. If there is a further problem, usually with fluctuating primary voltages, the taps may be adjusted to a different value. This is invariably a compromise.
In regards to added impedance, a 2% transformer and a 3% transformer will usually be set on the same taps and deliver the same open circuit voltage. If a 2% transformer is used and the impedance seen from the main breaker is increased to 3% either by the feeder impedance, an added reactor or a combination of the two, the effect will be the same as if a 3% transformer was used and there was no feeder impedance.
Respectfully