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 Post subject: Fault CurrentPosted: Mon Aug 16, 2010 8:11 am

Joined: Thu Apr 02, 2009 4:33 pm
Posts: 29
Location: OH
Would someone explain to me how the fault current process works from the utility through the transformer. If there is a fault on the secondary side of a transformer, this looks like a big load to the utility? For instance, a 75kva transformer, 12470/208 volt, 1.48 % impedence will have 14,066 amps of fault current on the secondary of the TX terminals with 235 amps of primary flowing, is this correct? When it is stated how much primary fault current is the utility supplying? is that how many amps the utility can supply on the non faulted primary side? Since there is 235 amps flowing on the primary side, how come there is more fault current needed in the formulas to get the 14,066 amps on the secondary?(Using infinite primary amps in the formula) Hope this does not sound too stupid.

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 Post subject: Posted: Mon Aug 16, 2010 1:50 pm
 Sparks Level

Joined: Sun Dec 23, 2007 1:44 pm
Posts: 348
Location: Charlotte, NC
rayman37 wrote:
When it is stated how much primary fault current is the utility supplying? is that how many amps the utility can supply on the non faulted primary side? Since there is 235 amps flowing on the primary side, how come there is more fault current needed in the formulas to get the 14,066 amps on the secondary

Whether it is given on primary or secondary depends on who owns the transformer as it is usually given at the point of service. Don't understand what you mean about the "more fault current needed in the formulas" though.

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 Post subject: Posted: Mon Aug 16, 2010 3:24 pm

Joined: Thu Apr 02, 2009 4:33 pm
Posts: 29
Location: OH
In the above problem, if the secondary is faulted, there will be 14,066 amps at the transformers secondary and 235 amps flowing on the primary side right? (using infinite utility current)? The 235 amps is the current using ratios right? If I plug the 235 amps in the formula when primary amps are known the fault current on the secondary is about half. I am missing something in my thought process or something in my math. Could someone explain how fault current works through the transformer from the utility from a faulted condition? During a fault how much current does the utility see on the primary side 235 amps or thousands on the primary side? During normal operation this loaded transformer is rated at 208 amps secondary with 3.47 on the primary side.

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 Post subject: Posted: Mon Aug 16, 2010 3:30 pm

Joined: Thu Apr 02, 2009 4:33 pm
Posts: 29
Location: OH
If someone is willing, I can call someone to help explain it to me. I know what I am trying to ask but I am having trouble trying to explain it if that makes any sense.

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 Post subject: Posted: Mon Aug 16, 2010 7:51 pm
 Arc Level

Joined: Wed Jun 04, 2008 9:17 am
Posts: 428
Location: Spartanburg, South Carolina
rayman37 wrote:
In the above problem, if the secondary is faulted, there will be 14,066 amps at the transformers secondary and 235 amps flowing on the primary side right? (using infinite utility current)? The 235 amps is the current using ratios right? If I plug the 235 amps in the formula when primary amps are known the fault current on the secondary is about half. I am missing something in my thought process or something in my math. Could someone explain how fault current works through the transformer from the utility from a faulted condition? During a fault how much current does the utility see on the primary side 235 amps or thousands on the primary side? During normal operation this loaded transformer is rated at 208 amps secondary with 3.47 on the primary side.

You need to distinguish between a fault on the primary and the current in the primary during a fault on the secondary. A fault on the primary is not restricted by the transformer impedance, but a fault on the secondary is.

Calculating 14,066A on the secondary and 235A on the primary is for a fault on the secondary side if the primary system (ahead of the transformer) had no impedance.

With this system, if you had a fault on the primary side of the transformer, the current would be infinite because there is no impedance to restrict it, thus the term "infinite bus".

I'm not sure what you mean by "plug the 235 amps in the formula when primary amps are known". If the impedance in the primary system was the same as that of the transformer, then a primary side fault would be 235A and a fault on the secondary would have a total impedance of twice that of the transformer alone. The current for a fault on the secondary would be 1/2 of what it was with an "infinite bus".

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 Post subject: Posted: Mon Aug 16, 2010 8:29 pm

Joined: Thu Apr 02, 2009 4:33 pm
Posts: 29
Location: OH
I guess what is confusing me is when using the ArcCalc software SKM, if I use infinite buss primary side the software calculates 14,066 amps secondary fault with 235 amps on the primary. If I use the Bussman software, when primary fault current amps are known, I have to enter 5,885 amps on the primary side to get 14,066 amps on the secondary side. Why are they different or am I missing something in the calculations?

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 Post subject: Posted: Tue Aug 17, 2010 5:23 am
 Arc Level

Joined: Wed Jun 04, 2008 9:17 am
Posts: 428
Location: Spartanburg, South Carolina
rayman37 wrote:
I guess what is confusing me is when using the ArcCalc software SKM, if I use infinite buss primary side the software calculates 14,066 amps secondary fault with 235 amps on the primary. If I use the Bussman software, when primary fault current amps are known, I have to enter 5,885 amps on the primary side to get 14,066 amps on the secondary side. Why are they different or am I missing something in the calculations?

Using "infinite bus" with SKM is the same as entering a very high primary fault current with Bussman. Anything over 5,885 amps will give the same results (neglecting fractional amps) as "infinite" because the transformer impedance is large with respect to the primary system impedance.

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 Post subject: Posted: Tue Aug 17, 2010 5:47 am

Joined: Thu Apr 02, 2009 4:33 pm
Posts: 29
Location: OH
So if I am understanding this correctly, there should only be 235 amps maximum on the primary side flowing when the secondary fault on the Transformer occurs in the above problem? Why do most programs want the available primary fault amps and why does that number need to be large for the same results vs the 235 amps flowing on the primary side?

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 Post subject: Posted: Tue Aug 17, 2010 6:16 am
 Sparks Level

Joined: Tue Feb 24, 2009 7:05 am
Posts: 252
Because if your source (what pushes amps to the primary side of the transformer) is too weak (ie, cannot push too many amps into the primary, what in electrical terms we call higher impedance), then for the same load it sees (a fault on the secondary of your transformer), less amps will flow than if it's a very strong source (lower or no impedance).

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 Post subject: Posted: Tue Aug 17, 2010 6:33 am

Joined: Thu Apr 02, 2009 4:33 pm
Posts: 29
Location: OH
I got it! Thanks Vincent and to all who answered.

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 Post subject: Posted: Tue Aug 17, 2010 7:10 pm

Joined: Thu Apr 02, 2009 4:33 pm
Posts: 29
Location: OH
One last question to make sure I understand things. I believe I was mixing two separate issues before. If there is a fault on the primary side of a transformer, the primary side will take what ever the utility (fault current) has to offer and pass this to the secondary side (scenario 1). If there is a fault on the secondary side of the transformer (scenario 2) this will look like a possible small load to the utility, but at rated voltage on the primary side, it will deliver thousands of amps on the secondary side and the primary side amperage will be whatever the ratio would be (transformation) like 235 amps or around that on the primary side using the above problem. Is my thinking right? Sorry to keep beating you guys up with questions.

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 Post subject: Posted: Tue Aug 17, 2010 8:16 pm
 Arc Level

Joined: Wed Jun 04, 2008 9:17 am
Posts: 428
Location: Spartanburg, South Carolina
rayman37 wrote:
One last question to make sure I understand things. I believe I was mixing two separate issues before. If there is a fault on the primary side of a transformer, the primary side will take what ever the utility (fault current) has to offer and pass this to the secondary side (scenario 1).

If the fault is on the primary side, there will be no current on the secondary side unless there is a source on the secondary, like a generator.
Quote:
If there is a fault on the secondary side of the transformer (scenario 2) this will look like a possible small load to the utility, but at rated voltage on the primary side, it will deliver thousands of amps on the secondary side and the primary side amperage will be whatever the ratio would be (transformation) like 235 amps or around that on the primary side using the above problem. Is my thinking right? Sorry to keep beating you guys up with questions.

More or less correct. It's more than a small load to the utility. If it is 67 time full load on the secondary, it is also 67 times full load on the primary. The utility will have protection to trip the transformer off-line if the secondary protection doesn't.

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 Post subject: Posted: Wed Aug 18, 2010 3:26 am

Joined: Thu Apr 02, 2009 4:33 pm
Posts: 29
Location: OH
Thanks again!

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