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Thread: Calculating available fault current.

  1. #1
    Join Date
    Sep 2009
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    Calculating available fault current.

    I've read through a lot of old posts and can't find exactly what I'm looking for.
    I am installing a 2500 Amp three phase 120/208 Volt service for a new apartment complex in New Haven. The local utility (UI) is asking me to show that they will have no more than 10,000 AIC at their meter cans. They told me that I had the possibility of 200,000 AIC at the load side of their transformer that is located in their vault right outside the building. I spoke with my engineer and he had no idea what they were looking for. I called a second engineer and he did a calculation but he came up over 10,000 AIC because the software he used would not allow he to input all the variables in the service. He said he didn't have the software needed or the training in the better software to input all the information. I'm in a bind. If anyone can point me in the right direction I would appreciate it. I guess what I need at this point is a recommendation of a person or company who has this software and training. Or let me know where I can get the software for a reasonable price to try to do this myself. But I don't know if I would have to be an Engineer to be able to present this calculation to the local utility.
    Thanks for your help.

  2. #2
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    Chapel Hill, NC
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    You have to calculate the fault current. Usually the poco does it to the transformer. Here is a free fault current calculator from Mike Holt. Plug in the numbers- you need excel.

  3. #3
    Join Date
    Sep 2009
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    9
    I appreciate the reply. I tried that before. I just tried it again and I came up with 22,538 Amperes at the panel/Meter. It's a great free program but I can't input enough info. I would need to be able to calculate for two Ferrez Shawmut CP-6 Fuses that protect the cable in case of a short that the utility is making me install on each side of every feeder. I'm hoping with these added it will bring me down under 10,000 at the meter. Here is a link to the fuse.

    http://us.ferrazshawmut.com/catalog/...tor-for-cu-al/

  4. #4
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    Alabama
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    If the utility is so concerned about their meter socket, use ct's. I have never heard of a requirement like this.

  5. #5
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    Illinois
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    You can't use current limiting fuses to reduce the available fault current at the meter can. The devices linked to are cable limiters and normally used to protect the legs of a parallel system to prevent a total outage when one of the parallel cables has a fault.

    There is not an easy way to limit the current at the meters on that system. You could use a reactor or really long service conductors,
    Don, Illinois
    "It is the first responsibility of every citizen to question authority." B Franklin

  6. #6
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    Illinois
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    Quote Originally Posted by dan1973ct View Post
    ... I would need to be able to calculate for two Ferrez Shawmut CP-6 Fuses that protect the cable in case of a short that the utility is making me install on each side of every feeder. ...
    These are normally only used at each end of a cable when the cables are installed in parallel. If there are multiple feeds from the transformer to the building that are not connected in parallel, they would normally only be used on the supply end.

    Is there any chance you can post a "one line" of this installation?
    Don, Illinois
    "It is the first responsibility of every citizen to question authority." B Franklin

  7. #7
    Join Date
    Mar 2004
    Location
    Collier County (Naples), Florida, USA
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    55

    Cool

    Mike H also has a fault current calculations book in his free stuff that I learned from to get the original excel file corrected to the present 7.1 version.
    http://www.mikeholt.com/documents/fr...ultcurrent.zip
    David A Engelhart
    Florida Electrical Inspector BN4045 Plans Examiner PX2057
    ICC (International Codes Council) Certified 1&2 Family Dwelling Inspector
    Florida Fire College and ICC Fire Safety Inspector I and Plans Examiner
    2011 President Joseph A Schneeberger/Florida Gulf Coast Division, Florida Chapter, Southern Section International Association of Electrical Inspectors (IAEI).

  8. #8
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    Wisconsin
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    Quote Originally Posted by dan1973ct View Post
    I would need to be able to calculate for two Ferrez Shawmut CP-6 Fuses that protect the cable in case of a short that the utility is making me install on each side of every feeder. I'm hoping with these added it will bring me down under 10,000 at the meter. Here is a link to the fuse.

    As Don pointed out, these protectors are not normally used for fault current reduction at equipment.

    However, you may want to confirm their performance with Mersen (formally Ferraz) now that you know the available fault current is only 22,538A. My guess is that they will not offer you any relief.
    Just because you can, doesn't mean you should.

  9. #9
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    Jul 2003
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    MA
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    Quote Originally Posted by dan1973ct View Post
    I've read through a lot of old posts and can't find exactly what I'm looking for.
    I am installing a 2500 Amp three phase 120/208 Volt service for a new apartment complex in New Haven. The local utility (UI) is asking me to show that they will have no more than 10,000 AIC at their meter cans. They told me that I had the possibility of 200,000 AIC at the load side of their transformer that is located in their vault right outside the building. I spoke with my engineer and he had no idea what they were looking for. I called a second engineer and he did a calculation but he came up over 10,000 AIC because the software he used would not allow he to input all the variables in the service. He said he didn't have the software needed or the training in the better software to input all the information. I'm in a bind. If anyone can point me in the right direction I would appreciate it. I guess what I need at this point is a recommendation of a person or company who has this software and training. Or let me know where I can get the software for a reasonable price to try to do this myself. But I don't know if I would have to be an Engineer to be able to present this calculation to the local utility.
    Thanks for your help.
    When you say at their meter cans, do you mean that the feeders go from the transformer through a distribution board and then to the individual meters?

    If you have a riser drawing that you can post, that shows feeder sizes, lengths and conduit types (PVC or Steel), I would like to try to help you with the calculation.

    200,000 AIC at the transformer seems high.
    That would be a 1000kva transformer with an impedence of 1.54% at 208v 3Ø (if my calculation is correct).
    Do you have any transformer information of just the AIC that UI gave you?
    Tim
    Master Electrician
    New England
    Yesterday's Technology at Tomorrow's Prices

    Answers based on 2011 NEC

  10. #10
    Join Date
    Jul 2006
    Location
    SE USA as far as you can go
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    They say 200kAIC on the load side of the transformer? That means you have either a transformer way bigger then needed, or a transformer impedance that is unlikely small. In that regard, I believe there are some questions which you can ask the utility that will help you figure this out.

    1. Verify the 200kAIC is the number they want you to use as the "Available short Circuit Current" on the 208V side of the utility transformer.

    2. What is the size of the transformer for your service in KVA or MVA, and the impedance of the transformer. This is going to tell you the maximum amount of fault current available for your installation. Also, get the HV and LV side rated voltages.

    (Keep in mind the answer from No.1 above may not match what you obtain by calculation from No.2. If that's the case, it could be they are giving you the ultimate long-term worst case)

    Add your cable into the mix. This should drop your current quite a bit, as the impedance of the cable will act like a choke.

    The MVA Method of calculating fault current works quite well for this situation; following is an example:

    Isc = 200KA @ 208V, 3 phase. Therefore you have:

    MVAsc = 1.732*208*200KA = 72MVAsc

    Now determine your cable MVA rating. Assume (6) 750kcmil per phase and the run is 200ft;
    MVAcble = (208)^2/((0.048/1000)*200) = 4.5MVA / ph and 4.5MVA*6 = 27MVAcble

    Series MVA combine like resistance in parallel, so;

    (72MVA*27MVA)/72MVA+27MVA) = 19.64MVA @ the service

    Isc=19.64MVA/(208*1.732) = 54.5kA

    Calculating backwards, you would need somewhere in the vicinity of a 1400ft run of (6) 750KCMIL per phase to get you down to 10KAIC. Not practical.
    All in all, I'd say the numbers don't add up. Additional clarification is needed from utility.
    "Just because you're paranoid, doesn't mean they're not out to get you"

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