autotransformer secondary fault currents

[basicbill]

Hi all...first post.
I am a journeyman electrician and I teach electrical apprentices. I have a question on how to calculate auto-transformer secondary fault currents.
I don't require engineering level calculations...just the concepts behind the calculation.

As I understand it an auto-transformer will 'transform' only a portion of the required secondary current and the rest is supplied directly from the primary line. Of course, there is no electrical isolation primary/secondary.

With this in mind, are secondary 'bolted fault' calculations able to be performed using the auto-transformer's kVA and %Z or rather should it be calculated using the previous transformer kVA rating and %Z? I realize that there is a choking effect of the series portion of the auto-transformer, line impedences etc.

But my main question is....would you use the ratings of the auto-transformer or the upstream isolating coil type transformer to perform fault current calculation.

I hope I have stated my question clearly.

Thanks in advance.

 

[davidbeach]

You really need to account for both. For fault calculations you can ignore the "auto" and just focus on the "transformer" part. Size(rating) and %Z are what you need.

 

[inpran]

@basicbill: Can you please explain what is a bolted fault?

 

[dpc]

A fault with zero impedance.

David Castor

http://www.cvoes.com

 

[basicbill]

Correct....thanks David.

So to re-state my question....

Do I have to consider the VA rating of the upstream transformer that supplies the auto-transformer when doing the fault current calculation? Since there is a direct electrical connection to it; as well as to the auto-transformer.

 

[basicbill]

As well as the %Z of the upstream transformer.

 

[ovide]

Hi BasicBill: The answer is " it depends how accurate an answer you want". I'll do an example : a Utility feeds a transfo (T1) rated 1000kva/ 480v-3ph secondary and z=5%. Assume the utility is infinite (an approximation). The 3ph fault at secondary of T1 is 1000kva/(.48kv*1.732*.05)=24ka .
This T1 feeds autotransfo T2,rated 225kva/440v-3ph sec and z=5%. ( another approx is to assume negligible z for the cable between T1 and T2)
a) if we ignore the upstream T1, then the 3ph fault on the sec of T2 (440v) = 225kva/(.44kv*1.732*.05)= 5.9ka .

b) if we include the effect of T1,we have to convert T1's impedance on T2's base and then add it to T2's z; so, this is done as follows : 24ka*.48kv*1.732 = 20mva ; .225mva/20mva = 0.01125 = 1.125% ( T1's equivalent z, as seen by T2 ); then the 3ph fault on the sec of T2 (440v) = 225kva/ (.44kv*1.732*.06125)= 4.82ka .
I hope this helps. ( another approximation was ignoring resistance component of z ). Ovide.