Calculating available fault current.

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dan1973ct

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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.
 
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...-protector/cp-600v-cable-protector-for-cu-al/
 
If the utility is so concerned about their meter socket, use ct's. I have never heard of a requirement like this.
 
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,
 
dan1973ct said:
... 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. ...
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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?
 
dan1973ct said:
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...-protector/cp-600v-cable-protector-for-cu-al/
Click to expand...


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.
 
dan1973ct said:
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.
Click to expand...

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 3? (if my calculation is correct).
Do you have any transformer information of just the AIC that UI gave you?
 
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.
 
It is possible that the utility is feeding this service with paralleled transformers, in which case 200kA is extremely realistic.
 
jim dungar said:
It is possible that the utility is feeding this service with paralleled transformers, in which case 200kA is extremely realistic.
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I think if the utility tells you that the available fault current if 200kA then you need to start from there.
You never know what their system consists of, or what they have planned in the future.
 
Thanks again for all the reply's. I will work on a hand drawing of a one line diagram and post it. I verified with the utility three times. They are adamant about 200,000. They are using two 500 KVA transformers in the vault. Here is a copy of what my engineer drew for the gear layout.
 

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I don't think the square d solution will help. If I understand the post correctly the equipment was sizes to handle the Arc fault current. I have to accommodate for the local utilities meters that are rated for 10,000.

I am attaching a drawing that lays out the distances that I was asked for earlier.
 

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dan1973ct said:
Thanks again for all the reply's. I will work on a hand drawing of a one line diagram and post it. I verified with the utility three times. They are adamant about 200,000. They are using two 500 KVA transformers in the vault. Here is a copy of what my engineer drew for the gear layout.
Click to expand...
OK. Maybe I'm missing something in translation here.

A 500 kVA transformer at 208V line to line would have a rated line current of about 1388A. For two in parallel rated current would then be around 2.8kA. At 5% impedance, short circuit current would then be about 56kA. Assuming zero impedance in the supply to the transformer. The 200kA doesn't seem credible to me.

Like I said, maybe I'm missing something.
 
Here is another view of the layout. This is a larger size. Hopefully this works.
As far as 200,000 AIC not being realistic, I don't have a say. UI told me I had to use 200,000 and work down to 10,000. At one point UI told my rep from Eaton that I could request the exact AIC and when I did it took them almost a week to get back to me and tell me I had to use 200,000 at the load side of their transformer.
 

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ABB makes a device called an Is-limiter, works well for this situation. You can also use a current limiting reactor, essentially a 1:1 transformer with the proper reactance to choke the fault current.

You can also use them in parallel if you can't afford to take the chance of a low voltage condition during normal operation, i.e. non-fault. Here's a link to help you get started, and yes, this stuff does work.

http://www.abb.com/product/db0003db004279/c125739900636470c125698c00553a43.aspx?tabKey=6
 
Besoeker said:
OK. Maybe I'm missing something in translation here.

A 500 kVA transformer at 208V line to line would have a rated line current of about 1388A. For two in parallel rated current would then be around 2.8kA. At 5% impedance, short circuit current would then be about 56kA. Assuming zero impedance in the supply to the transformer. The 200kA doesn't seem credible to me.

Like I said, maybe I'm missing something.
Click to expand...

jim dungar said:
It is possible that the utility is feeding this service with paralleled transformers, in which case 200kA is extremely realistic.
Click to expand...

Besoeker, You may be "misssing" the possibility that the transformers are part of a POCO grid as Jim notes. In our larger cities it is fairly common to have 200ka + where the POCO grid consists of several transformers paralleled
 
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