Convert secondary fault current to the primary

[mikefinn]

thread238-163018

In reference to Thread 238-163018.

I am having trouble converting a secondary fault current to the primary of a transformer.

I believe you can use either the PU method or the MVA method, however my results are coming out too small.

Here the numbers I am using if it helps:
1500kVA
Vpri = 24,900V, Vsec = 480V, %Z=5.6%
Isc (sec) = 32kA

Thanks, Mike

 

[burnt2x]

I_pri = 1,500,000/(1.732 x 24,900) = 34.78A .
I_prisc= 34.78/0.056pu = 621 A

 

[mikefinn]

Thanks for commenting.
I see you took the FLAs on the primary and divided it by the impedance of the XFMR to get the fault current on the primary. By doing this don't you only take into account the XFMR??

What if there was other contribution from motors or generators?

I am looking for a way/formula to convert a given fault current on the secondary (32kA) to the primary while taking into account the impedance of the XFMR.

thanks.

 

[stevenal]

Multiply your 32kA by 480/24900.
Other contributions from the secondary side for a secondary fault do not contribute to winding current.

 

[mikefinn]

I agree, however that would not take into account the impedance of the XFMR.

 

[stevenal]

You used the transformer impedance to get the 32kA. Impedance only impedes once.

 

[davidbeach]

What are you asking?

Are you asking how many amps will there be on the primary for a 32kA secondary fault? Or, are you asking how much fault current would there be for a primary fault if there is 32kA for a secondary fault?

Two very different questions. The first is well answered already, but from the sound of your follow up questions you might be asking the second.

 

[mikefinn]

Yes, I was given an available fault current at the secondary of the XFMR (by applying a fault at the secondary). I am looking for the available fault current on the primary (by applying a fault at the primary of the XFMR).

So the second question, a primary fault if there is 32kA on the secondary. Thanks.

 

[Zogzog]

The available fault current on the primary is dependant on it's source

 

[davidbeach]

So, what you need to do is set up a equivalent system model consisting of an infinite bus, a source impedance, and the transformer impedance. Given 32kA fault current, solve for the source impedance. Then remove the transformer impedance and solve for the fault current.

 

[tem1234]

If you take an infinite source with your transformer, the SC at the txfo secondary will be:

1500kVA/1.73/480V/5.6% = 32 218 A

So for a 32 kA SC at the secondary, the source is almost infinite, so the SC level at 24.9kV is infinite.

People often do this assumption, it's a conservative way for sizing a breaker at the secondary and you don't have to ask the utility SC current. If you want the real SC current at the primary, ask the utility(or anybody who have this info). With this information, you will be able to calculate a more realistic SC at the secondary too.

 

[PHovnanian]

If you want the real SC current at the primary, ask the utility(or anybody who have this info).

If they'll give you the figure, that is. Utilities are free to change their system configurations. They don't want to become liable should a customer size equipment based on an outdated figure and then suffer excessive damage following a fault. So the number you'll get will be very conservative.

 

[burnt2x]

For a fault at the point where the transformer is fed from the system, the only impedance in between the fault point and the system is the system impedance. As pointed out by tem1234, if the source/system is infinite(zero impedance), the fault current will be "infinitely large"!

 

[davidbeach]

True, but then the fault current on the secondary isn't really 32kA. If that is the case (I've not bothered with the math) then the 32kA value is strictly for equipment rating and has nothing to do with actual fault values. Do not use this value for arc-flash studies.

 

[burnt2x]

Yup. My first post was based on the understanding that the fault was at the secondary side of the trafo. If the fault occurs at the primary side, the fault current depends on the source impedance.

 

[erickench]

First of all you would have to determine the turns ratio. In this case it would be:

24,900/480=51.875 say 52

If the the short circuit current on the secondary is 32,000A then the primary Isc would be 32,000/52=615.4A

Eric Kench, P.E.

 

[davidbeach]

No, that would be the correct answer to "Are you asking how many amps will there be on the primary for a 32kA secondary fault?" but it is not the correct answer to "Or, are you asking how much fault current would there be for a primary fault if there is 32kA for a secondary fault?"

The question being asked is "Or, are you asking how much fault current would there be for a primary fault if there is 32kA for a secondary fault?" and the answer is that there is insufficient information to answer the question because 32kA is a nice round number approximately what you get if you have an infinite bus on the high side of the transformer.

Helps to read the whole thread before answering.

 

[mikefinn]

Thank you to everyone who commented. Much appreciated.

Mike